Category: CBSE Class 10

NCERT Solutions for Class 10 Science Chapter 15 Our Environment

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 15 Our Environment. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 15 – Our Environment solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 15 – Our Environment Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
Why are some substances biodegradable and some non-biodegradable ? (CCE 2011)
Answer:
Biodegradable substances are substances of organic origin which can be broken down by enzymes of saprophytes, e.g., garbage, sewage, livestock waste, agriculture waste. Non-biodegradable substances are man-made substances which cannot be degraded by saprophytes because they lack the enzymes to do so e.g., waste plastic and polythene articles.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2..
Give any two ways in which biodegradable substances would affect the environment.
(CBSE Delhi 2007, AI 2009 C)
Answer:

1. Stink: Within a day or so waste biodegradable substances begin to stink and produce foul gases.
2. Pests and Pathogens: The decaying biodegradable substances become breeding places of flies and many other pests. They also contain a number of pathogens. Flies and other pests carry the germs to all the places visited by them resulting in spread of diseases.

Question 3.
Give any two ways in which non-biodegradable substances would affect the environment. (CBSE AI 2009 C)
Answer:

1. Dumping Area: Dumping of non-biodegradable substances on a piece of land converts the same into barren land. It is also called landscape pollution.
2. Biological Magnification: Pesticides, heavy metals and other chemicals enter water and food chains. They accumulate in toxic proportions and harm all kinds of living organisms. Their concentration also increases with rise in trophic level. Human beings are harmed the most because man lies at the top of every food chain.

Question 4.
What are trophic levels ? Give an example of a food chain and state the trophic levels in it.
Answer:
(a) Trophic Levels. They are steps or divisions of food chain which are characterised by particular methods of obtaining food, e.g., producers (T₁), herbivores (T₂), primary carnivores (T₃), etc.

(b) Flow of Energy Through Food Chain. Energy enters a food chain through producers. Producers or green plants trap solar energy and convert it into chemical energy of food during photosynthesis. From producers energy passes into herbivores. A lot of energy dissipates during transfer and utilization of food energy by herbivores (10% law). From herbivores the food energy passes into primary carnivores, again with a lot of dissipation. Only about 10% of herbivore energy is passed into body mass of primary carnivores. From primary carnivores, nearly 10% energy passes into secondary carnivores and so on. It is ultimately lost as heat.

Question 5.
What is the role of decomposers in the ecosystem ?
Answer:
Ecosystem is a self-contained ecological system which consists of a distinct biotic community and the physical environment, both interacting and exchanging materials between them. It is a structural and functional unit of biosphere. Ecosystem is an open system which requires a regular input of energy and circulation of matter for its sustenance. Different ecosystems are in contact with one another. Though they are distinct, but exchange of materials can occur amongst them. Ecosystem can be small (e.g., rotting piece of wood) or large (e.g., forest, ocean), temporary (e.g., rain fed pond) or permanent (e.g., lake, forest), natural or artificial.
Natural Ecosystems: They are ecosystems which develop in nature without human support. Natural ecosystems are of two types, terrestrial and aquatic. Terrestrial ecosystems occur over land. They are of three major types -— desert, grassland and forest. Aquatic ecosystems are found in water bodies, e.g., ponds, lakes, rivers (fresh water), estuaries, marine (salt water).
Artificial Ecosystems: They are ecosystems which have been p/g. 5.7. An aquarium. created and are maintained by human beings. Artificial ecosystems are also called man-made or anthropogenic ecosystems. Agroecosystem is the largest man-made ecosystem. Garden is a common artificial ecosystem maintained by most institutes. It has various types of plants grown and maintained by gardener — grass, trees, flower bearing plants like Rose, Jasmine, Sunflower. A number of animals become residents and visitors of the garden. Aquarium is another artificial ecosystem.

Question 6.
What is ozone and how does it affect any ecosystem ?
Answer:
Definition of Ozone: Ozone is a triatomic molecule made up of three atoms of oxygen, O₃.
Effect:

1. Protection against ultraVoilet rays if present in stratosphere,
2. Ozone dissipates the energy of ultraviolet rays by undergoing dissociation following by reassociation
   

If present in the atmosphere of ecosystem, it is highly toxic causing injury to mucous membranes, eye irritation and internal haemorrhages in animals and humans. It harms plants by destroying photosynthetic cells producing necrosis.

Question 7.
How can you help in reducing the problem of waste disposal ? Give any tivo methods.
Answer:
Waste disposal is becoming the biggest problem of all local bodies. It is consuming a big chunk of their revenues. Even then the points of collection are turning into stinking areas. Transport of waste to disposal site is seldom perfect. There is stink all the way. A lot of waste is thrown on the roads by the vehicles carrying the waste.
The quantity of waste can be reduced if you form the habit of separation of waste into biodegradable and non-biodegradable parts.

1. Non-biodegradable Waste. Most often the non-biodegradable waste is recycled. It is taken away by rag pickers.
2. Biodegradable Waste. Biodegradable waste is putressible. It can be composted or vermicomposted to prepare compost for your kitchen gardens. The technique can also be used in schools and other institutions. Some institutes in big cities have installed incinerators for disposal of combustible components of waste. Hospital waste is being compulsorily incinerated to reduce contamination and spread of diseases.

NCERT Chapter End Exercises

Question 1.
Which of the following groups contains only easily biodegradable items ?
(A) Grass, flower and leather
(B) Grass, wood and plastic
(C) Fruit peels, cake and lime juice ,
(D) Cake, wood and grass.
Answer:
(A), (C), (D).

Question 2.
Which of the following constitute a food chain
(A) Grass, wheat and mango
(B) Grass, goat and human
(C) Goat, cow and elephant
(D) Grass, fish and goat.
Answer:
(B).

Question 3.
Which of the following are environment friendly practices ?
(A) Carrying cloth bags to put purchases in while shopping
(B) Switching off unnecessary lights and fans
(C) Walking to school instead of getting your mother to drop you on her scooter
(D) All the above.
Answer:
(D).

Question 4.
What will happen if you kill all the orgainsms in one trophic level ?
Answer:
Higher Trophic Level. All the organisms will be starved and die. Lower Trophic Level. The organisms will increase in number much beyond the carrying capacity of the enviroment. They will become weak and fall easy prey to various diseases, ultimately resulting in decline in population. The ecosystem may get converted into a desert.

Question 5.

1. Will the impact of removing all the organisms in a trophic level be different for different trophic levels ?
2. Can the organisms of any trophic level be removed without causing any damage to the ecosystem ?

Answer:

1. Yes. Impact of removing all the organisms of a trophic level depends upon the trophic level. Removal of producers will kill all the consumers. Killing of carnivores will increase the number of herbivores which in turn will eat up all the producers. They would themselves die of starvation. However, if herbivores are removed, all the carnivores of the area will die. The plants which are dependent upon animals for pollination will fail to reproduce and die ultimately. Others would persist and increase in number, again much beyond the carrying capacity of the environment.
2. No. Removal of all the organisms of a trophic level will disturb the ecosystem – killing of higher trophic level organisms and explosion in populations of lower level organisms. Higher number of lower trophic level organisms will adversely affect the ecosystem by consuming whole or major part of their prey.

Question 6.
What is biological magnification ? Will the levels of this magnification be different at different levels of the ecosystem ?
Answer:
Definition: Biological magnification is increase in the concentration of a chemical per unit weight of the organisms with the successive rise in trophic level. In one study it was found that concentration of harmful chemical like DDT will increase 80,000 times the concentration present in water.
Levels of Magnification at Different Levels of Ecosystem. Levels of biomagnification of a chemical will be different at different levels of ecosystem. In one study it was found that concentration of a pesticide was 0.002 ppm in water, 0.05 ppm in plankton, 2.4 ppm in fish and 16.0 ppm in fish eating bird.

Question 7.
What are the problems caused by the nonbiodegradable waste we generate ?
Answer:

1. Volume: Being non-biodegradable, the volume of these wastes will not decrease by natural process of decomposition.
2. Land Use: They have to be dumped over land. The land becomes unfit for any other purpose.
3. Heavy Metals: Heavy metals present in industrial wastes (e.g., copper, lead, chromium, nickel, mercury) remain in the soil indefinitely. Slowly they pass into vegetation and crops harming both humans and animals.
4. Pesticides and Other Toxins: They pollute underground water, surface water and soil. The chemicals enter food chain and become concentrated, harming animals and humans.
5. Rag Pickers: In the process of removing recyclable materials from solid wastes, the rag pickers are exposed to many diseases and toxins.
6. Recycling: Recycling of materials produces only’inferior quality products. Recycling of polythene and plastic is accompanied by release of carcinogenic toxins like dioxins and furans.

Question 8.
If all the waste we generate is biodegradable, will this have no impact on the environment ?
Answer:
The impact on the environment will depend upon the system of collecting, transporting and disposal of biodegradable waste. If it is regular and clean, the impact will be little in urban areas. Only some effect will be observed at disposable site which will emit foul gases and expose the sanitary workers to contamination. The same can also be reduced by increased use of machines and wearing of protective gear by the sanitary workers.

Question 9.
Why is damage to the ozone layer a cause of concern What steps are being taken to limit this damage ?
Answer:
Cause of Concern: Ozone layer present in the stratosphere has thinned out by about 8% over the equator and more so over the antarctica where a big ozone hole appears every year.
This has increased the level of UV-B radiations reaching the earth by 15-20%. These radiations are causing increased number of skin cancers, cataracts and reduced immunity in human beings. There is increased incidence of blinding of animals, death of young ones, reduced photosynthesis, higher number of mutations and damage to articles.
Steps to Limit Damage:

1. Ban on production and use of halons.
2. Ban on production and use of chlorofluorocarbons.

Selection Type Questions

Alternate Response Type Questions
(True/False, Right(√)/Wrong (x), Yes/No)

Question 1.
Wastes are of two types, biodegradable and non-biodegradable.
Question 2.
Non-biodegradable articles are the ones which cannot be digested.
Question 3.
Ozone is formed in stratosphere by action of ultravoilet radiations on oxygen.
Question 4.
Earth is kept warm due to green house flux.
Question 5.
Rag pickers remove reusable articles.
Question 6.
Pyrolysis is aerobic combustion while incineration is anaerobic combustion.
Question 7.
Biodegradable wastes should be separated and kept in blue colour bins for garbage collectors.
Question 8.
Food web ensures survival of all types of trophic levels.

Matching Type Questions

Question 9.
Match the articles in columns I and II (single matching) :

Column I	Column II
(a) Hawk	(i) Producer
(b) Hare	(ii) Top carnivore
(c) Grass	(iii) Green house gas
(d) Carbon dioxide	(iv) Herbivore

Question 10.
Match the contents of columns I, II and III (double matching)

Column I	Column II	Column III
(a)    Ecosystem (b)    Producers (c)    Omnivores (d)    Persistent pesticide	(i) Food chain (ii) Abiotic components (iii) Transducer (iv) Plant food	p.  Food energy q.  Biomagnification r. Animal food s. Biotic components

Question 11.
To what trophic level (T₁, T₂, T₃, T₄) do the following belong (key or check list items)

Question 12.
Match each stimulus with appropriate response :

Activity	Deforestation (A)	Increase runoflf (B)	Destruction of wild life (C)
(i) Hunting (ii) Felling of trees (iii) Litter Collection

Fill In the Blanks

Question 13. Waste substances that are broken down by microbes are called ………. : ………… .
Question 14. Gardens and crop fields are ………………. ecosystems.
Question 15. Microorganisms are called ……………… as they break down the complex organic remains into simple inorganic substances.
Question 16. Decrease in ozone in stratosphere is linked to release of synthetic chemicals like ……………… .
Question 17. Improvement in life style often results in increased generation of …………. : ………… material.

Answers:

NCERT Solutions for Class 10 Science Chapter 15 Our Environment

Hope given NCERT Solutions for Class 10 Science Chapter 15 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 14 Sources of Energy

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 14 Sources of Energy. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 14 – Sources of Energy solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 14 – Sources of Energy Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
What is good source of energy ?
Answer:
A substance which produces sufficient amount of energy over a long period of time without producing much smoke is known as good source of energy. Characteristics of a good source of energy.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
What is good fuel ?
            Or
What is a fuel ?
Answer:
A fuel which provides large amount of heat energy without causing pollution.

Question 3.
If you could use any source of energy for heating your food, which one would you use and why ?
Answer:
We would use microwave oven for heating our food. This is because the nutritional value of food is not lost when heated in a microwave oven.

Question 4.
What are the disadvantages of fossil fuels ? (CBSE 2010)
Answer:

1. They cause environmental pollution.
2. They cause global warming.
3. They do not supply enough heat energy.
4. The by-products of burning fuels cause acid rain which pollute water resources.

Question 5.
Why are we looking at alternate sources of energy ?
                                                      Or
State one reason that has necessitated to look for alternate sources of energy ? (CBSE 2013)
Answer:
Because the conventional sources of energy may completely be exhausted one day if their use at the present rate continues.

Question 6.
How has the traditional use of wind energy and water been modified for our convenience
(CBSE 2010, Term I)
Answer:
These energies have been converted into electrical energy using electric generators.

Question 7.
What kind of mirror—concave, convex or plane would be best suited for the use in a solar cooker. Why ?
Answer:
Concave mirror, because it focusses the sun rays at a point to raise the temperature at that point.

Question 8.
What are the limitations of the energy that can be obtained from oceans ? (CBSE 2010)
                                                      Or
List any four limitations of the energy obtained from oceans. [CBSE (All India) 2009, 2012]
Answer:

1. Energy from oceans is available only when high tides are in the ocean.
2. Power plants used to convert ocean energy into electric energy do not operate continuously.
3. Energy of ocean waves can be extracted blow all times across the ocean.
4. Tidal power plants can not be installed everywhere.

Question 9.
What is geothermal energy ? (CBSE 2010, Term I, 2011, 2012)
Answer:
The heat energy stored in the hot spots of the earth’s crust is called geothermal energy.

Question 10.
What are the advantages of nuclear energy ?
Answer:

1. A small quantity of nuclear fuel gives a large amount of energy.
2. In a nuclear power plant, the nuclear fuel is inserted once to get energy over a longer period of time.

Question 11.
Can any source of energy be pollution free ? Why or why not ?
Answer:
No source of energy is there which is pollution free. However, some sources of energy cause more pollution and some sources of energy cause less pollution.

Question 12.
Hydrogen has been used as a rocket fuel. Would you consider it a cleaner fuel than CNG? Why or why not ?
Answer:
Hydrogen causes less air pollution than C.N.G. because burning of hydrogen produces water vapours and burning of CNG produces CO₂. When the concentration of CO₂ increases in the atmosphere, then the temperature of the atmosphere increases. This effect is known as green house effect. The increased temperature of the atmosphere affects life on the earth.

Question 13.
Name two energy sources that you would consider to be renewable. Give reasons for your choices.
Answer:

1. Bio-mass is considered as a renewable source of energy because forests can be replenished.
2. Water is also a renewable source of energy as water is continuously available to use due to water cycle in nature.

Question 14.
Give the names of two energy sources that you would consider to be exhaustible.
Answer:
Give reasons for your choices,

1. Coal
2. Petroleum.

They will be exhaustible when continuously extracted. Moreover, the formation of these fuels under the earth takes a longer period of time.

NCERT Chapter End Exercises

Question 1.
A solar heater cannot be used to get hot water on
(a) a sunny day,
(b) a cloudy day,
(c) a hot day,
(d) a windy day.
Answer:
(b).

Question 2.
Which of the following is not a example of a bio-mass energy source ?
(a) wood,
(b) gobar gas,
(c) fossil fuels,
(d) bio-mass. (Bihar Board 2012)
Answer:
(c).

Question 3.
Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the Suns energy ?
(a) geothermal energy,
(b) wind energy,
(c) fossil fuels,
(d) bio-mass.
Answer:
(a).

Question 4.
Compare and contrast fossil fuels and the Sun as a source of energy.
Answer:

1. Energy of fossil fuels comes from the solar energy. However, fossils fuels are the non-renewable sources of energy. On the other hand, sun is a renewable source of energy.
2. Fossil fuels cause pollution but solar energy does not cause pollution.
3. Energy is supplied by fossil fuels at any time of the day but sun supplies energy only when it shines.

Question 5.
Compare and contrast bio-mass and hydro-electricity as source of energy.
Answer:

Bio-mass	Hydro electricity
(i)    The energy supplied by the burning of bio-mass causes pollution. (ii)    The energy from bio-mass can be obtained by burning it directly or by a gobar gas plant.	(i)     Hydro-electricity does not cause pollution. (ii)    Hydro-electricity can be obtained by constructing costly dams.

Question 6.
What are the limitations of extracting energy from
(a) the wind
(b) Waves
(c) Tides ?
Answer:
(a)

1. We cannot depend upon wind energy as it is available only when strong wind blows. The appliances or machines operating with wind energy stop working as soon as wind stops. The minimum speed of wind to operate generator to produce electricity is 15 km/h.
2. Wind energy is not sufficient to operate heavy machines.
3. The use of wind energy is limited to certain places where strong winds blow most of the time.

(b) Energy of waves can be extracted only if strong winds blow all the time across the sea.
(c) Tidal power plant can extract energy from the waves only when the difference between the water levels of high tide and low tide is very large.

Question 7.
On what basis you classify energy source as
(a) Renewable and non-renewable ?
(b) Exhaustible and inexhaustible ?
(c) Are the options given in (a) and (b) the same ?
Answer:
(a) Renewable Sources of Energy are those which are continuously supplied by nature. For example, the sun, the wind.
(b) Non-Renewable Sources of Energy are those which have been formed in nature long ago under certain conditions of temperature and pressure. Non-renewable sources of energy take longer period of time for their formation. For example, fossil fuels like coal, petroleum.
(c) Inexhaustible Sources of Energy are those which supply continuous energy for unlimited time. In fact, exhaustible sources of energy are also termed as non-renewable energy sources. But wood is an exception as it can be made renewable by growing more plants periodically. Inexhaustible sources of energy are termed as renewable sources of energy.

Question 8.
What are qualities of an ideal source of energy ?
                                               Or
Write any two characteristics of a good source of energy. (CBSE 2013)
Answer:

1. It supplies useful energy continuously.
2. It does not cause environment pollution.
3. It is economical.

Question 9.
What are the advantages and disadvantages of using a solar cooker ? Are there places where solar cookers would have limited utility ? (CBSE 2010)
Answer:
Advantages:

1. There is no cost of cooking food in a solar cooker.
2. No pollution is caused when food is cooked in a solar cooker.
3. Nutrition value of food is preserved when food is cooked in the solar cooker.
4. Two or three dishes can be cooked at a time.

Disadvantages:

1. Food cannot be cooked at night and on a cloudy day using a solar cooker.
2. The cost of making solar cooker is high.
3. Food cannot be cooked quickly with the solar cooker.
4. Large quantity of food cannot be cooked with a solar cooker.
5. Chapatis cannot be made with solar cooker.
   Solar cooker will have limited utility at places where the sun shines for shorter period of time or where the sun rays never reach.

Question 10.
What are the environmental consequences of the increasing demand for energy ? What steps would you suggest to reduce energy consumption ?
Answer:

1. More use of fossil fuels for fulfilling the increasing demand for energy is polluting air (i.e. the environment),
2. LPG and CNG are considered as clean fuels but the extraction and transportation of these fuels cause environmental pollution.
3. The use of large number of sources of energy is causing global warming.

Suggestions:

1. Burning fuels must be extinguished as soon as their use is over.
2. Bulbs, tubes and other electrical appliances must be switched off as soon as you leave your room.
3. The engines of the vehicles must be switched off when these vehicles are stopped for more than a minute to save fuel.

NCERT Solutions for Class 10 Science Chapter 14 Sources of Energy

Hope given NCERT Solutions for Class 10 Science Chapter 14 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current. LearnInsta.com  provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 13 – Magnetic Effects of Electric Current solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 13 – Magnetic Effects of Electric Current Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
Why does a compass needle get deflected when brought near a bar magnet ?
Answer:
Compass needle is a small magnet which experiences a force in the magnetic field of a bar magnet. Due to this force, it gets deflected.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Draw magnetic field lines around a bar magnet. (CBSE 2011, 2012, 2013, 2014)
Answer:

Question 3.
List the properties of magnetic field lines. (CBSE 2015)
                                            Or
Write four properties of magnetic field lines. (CBSE (Delhi) 2010, 2011, 2012, 2013)
Answer:

1. Magnetic field lines are closed continuous curves.
2. The tangent at any point on the magnetic field line gives the direction of the magnetic field at that point.
3. No two magnetic field lines can cross each other.
4. Magnetic field lines are crowded in a region of strong magnetic field and magnetic field lines diverge in a region of weak magnetic field.

Question 4.
Why do not two magnetic field lines intersect each other ? (CBSE 2011, 2012, 2015)
                                         Or
No two magnetic field lines can intersect each other. Explain. (CBSE 2010, 2014)
Answer:
The tangent at any point on a magnetic field line gives the direction of magnetic field at that point. If two magnetic field lines cross each other, then at the point of intersection, there will be two tangents. Hence, there will be two directions of the magnetic field at the point of intersection. This is not possible. Hence, no two magnetic field lines can cross each other.

Question 5.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right hand rule to find out the direction of magnetic field inside and outside the loop.
(CBSE (Delhi) 2009)
Answer:
Magnetic field inside the loop is perpendicular to the plane of table and in the downward direction. However, outside the loop, magnetic field is perpendicular to the plane of the table and in the upward direction.

Question 6.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
(CBSE 2011, 2012, 2015)
Answer:

Question 7.
Choose the correct option :
The magnetic field inside a long straight solenoid carrying current is :
(a) zero
(b) decreases as we move towards its ends .
(c) increases as we move towards its ends
(d) is the same at all points.
Answer:
(b).

Question 8.
Which of the property a proton can change when it moves freely in a magnetic field ? (There may be more than one correct answer).
(a) mass
(b) speed
(c) velocity
(d) momentum.
Answer:
A force acts on a proton when it moves freely in a magnetic field. Hence its velocity and momentum can change.

Question 9.
In activity 13-7 (NCERT book), how do we think the displacement of rod AB will be affected if

1. current in rod AB is increased
2. a stronger horse shoe magnet is inserted
3. length of the rod AB is increased.

Answer:
Force acting on a current carrying conductor of length l placed perpendicular to magnetic field B is given by F = B I l

1. When I increase, F also increases. Hence the displacement of the rod increases.
2. When a stronger horse shoe magnet is inserted, magnetic field at B increases. So force also increases. Hence displacement increases.
3. When l increases, force increases and hence displacement increases.

Question 10.
A positively charged partical (alpha partical) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is
(a) towards south
(b) towards east
(c) downward
(d) upward.
Answer:
(d).

Question 11.
State Fleming’s left hand rule with a labelled diagram. (CBSE 2005, 2010, 2011, 2012)
Answer:
Fleming’s left hand rule
Statement : Stretch the left hand such that the thumb, first finger and the central finger are mutually perpendicular to each other. If the First finger points in the direction of the magnetic Field and the Central finger points in the direction of Current, then the thumb will point in the direction of Motion (or Force) as shown in figure 24.

Thus, in Fleming’s left hand rule, first finger shows the direction of the magnetic field. The central finger shows the direction of electric current flowing in the conductor. Thumb shows the direction of force on the conductor or the direction of motion of the conductor

Question 12.
What is the principle of electric motor ? (CBSE 2012)
Answer:
Electric motor works on the principle that a current carrying conductor placed perpendicular to a magnetic field experiences a force.

Question 13.
What is the role of the split ring in an electric motor ?
Answer:
The split-ring in an electric motor reverses the direction of current in the armature coil of the motor. Therefore, the direction of the force acting on the two arms of the coil is also reversed. As a result of this, the coil of d.c. motor continues to rotate in the same direction.

Question 14.
Explain different ways to induce current in a coil. (CBSE 2010, 2011, 2012)
Answer:

1. By moving a bar magnet towards or away from the coil.
2. By placing a closed coil near another coil connected across a battery.

Question 15.
State the principle of electric generator.
Answer:
It is based on the principle of electromagnetic induction. That is, the changing magnetic field induces current in the coil.

Question 16.
Name some sources of direct current. (Bihar Board 2012, 2015)
Answer:
A dry cell, a battery, a solar cell, d.c. generator etc. are some sources of direct current.

Question 17.
Which source produces alternating current ?
Answer:
AC generator (which converts mechanical energy into alternating current or electricity) and an oscillator (a device which converts D.C. into A.C.) are the sources which produce alternating current.

Question 18.
Choose the correct option :
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution,
Answer:
(c).

Question 19.
Name two safety measures commonly used in electric circuit and appliances.
(CBSE 2010, 2012, 2014, 2015)
Answer:

1. Electric fuse and
2. earthing.

Question 20.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect ? Explain. (CBSE 2010, 2011, 2012, 2013, 2014)
Answer:
P = 2 kW = 2000 W and V = 220 V

This shows that current flowing through the oven is more than the current rating (5 A). Hence, the fuse in the circuit melts and oven is saved from damage.

Question 21.
What precautions should be taken to avoid the overloading of domestic electric circuit ? (CBSE 2010, 2011, 2012, 2015)
                                                                         Or
Write any two precautions to be taken to avoid overloading of a domestic electric circuit.
Answer:

1.  We should not connect many appliances in the same socket.
2. Electrical appliances of high power rating should not be switched on simultaneously.

NCERT Chapter End Exercises

Question 1.
Which of the following correctly describes the magnetic field near a long straight wire ?
(a) the field consists of straight lines perpendicular to the wire
(b) the field consists of straight lines parallel to the wire
(c) the field consists of radial lines originating from the wire
(d) the field consists of concentric circles centered on the wire.
Answer:
(d).

Question 2.
The phenomena of electromagnetic induction is
(a) the process of charging a body
(b) the process of generating magnetic field due to current passing through a coil
(c) producing induced current in a coil due to relative motion between a magnet and the coil
(d) the process of rotating a coil of an electric motor.
Answer:
(c).

Question 3.
The device used for producing electric current is called a
(a) generator
(b) galvanometer
(c) ammeter
(d) motor. (Bihar Board 2012)
Answer:
(a).

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electromagnet while a DC generator has permanent magnet
(b) DC generator will generate a higher voltage
(c) AC generator will generate & higher voltage
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(d).

Question 5.
At the time of short circuit, the current in the ciruit
(a) reduces substantially
(b) does not change
(c) increases heavily
(d) vary continuously. (Bihar Board 2012)
Answer:
(c).

Question 6.
State whether the following statements are true or false :
(a) an electric motor converts mechanical energy into electrical energy ‘
(b) an electric generator works on the principle of electromagnetic induction
(c) the field at the centre of a long circular coil carrying current will be parallel straight lines
(d) a wire with a green insulation is usually the live wire.
Answer:
(a) False. It converts electrical energy into mechanical energy.
(b) True.
(c) True.
(d) False. Five wire has red insulation cover.

Question 7.
List three sources of magnetic fields. (CBSE 2013)
Answer:

1. a permanent magnet
2. a current carrying conductor
3. a current carrying solenoid.

Question 8.
How does a solenoid behave like a magnet ? Can you determine the north and south poles of a current carrying solenoid with the help of a bar magnet. Explain. (CBSE 2012, 2013)
Answer:
When electric current flows through a solenoid, magnetic field is set up around the solenoid. The pattern of the magnetic field is same as that of the magnetic field of a bar magnet. One end of the solenoid behaves as north pole and the other end of the solenoid behaves as south pole.
To determine the north and south poles of a current carrying solenoid with the help of a bar magnet, suspend it with a strong thread. Now bring the north pole of a bar magnet towards one end of the solenoid. If the solenoid attracts towards the magnet, then that face of the solenoid is south pole. If the bar magnet moves away from the solenoid, then that face of the solenoid is the north pole.

Question 9.
When is the force experienced by a current carrying conductor placed in magnetic field is the largest ?
[CBSE, (All India) 2009, 2010]
                                                                              Or
Under what condition does a current carrying conductor kept in a magnetic field experience maximum force ? (CBSE 2011, 2012, 2014, 2015)
Answer:
When current carrying conductor is placed perpendicular to the magnetic field.

Question 10.
Think you are sitting in a chamber with your back to one wall. An electron beam moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field ?
Answer:
Movement from electron beam from back wall to the front wall is equivalent to the flow of electric current from front wall to the back wall. The deflection of the beam means, the force is acting towards our right side. According to Fleming’s Left Hand Rule, the direction of magnetic field is vertically downward. That is, the magnetic field is perpendicular to the plane of the paper and directed inward.
Such magnetic field is shown by ⊗.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor ?
Answer:
Electric motor converts electrical energy into mechanical energy.
Principle: Electric Motor is based on the fact that a current carrying conductor placed perpendicular to the magnetic field experiences a force.

1. Armature coil: It consists of a single loop of an insulated copper wife in the form of a rectangle. Rectangle ABCD shown in figure 25 is an armature coil.
   
2. Strong field magnet: Armature coil is placed between two pole pieces (N and S poles) of a strong magnet. This magnet provides a strong magnetic field.
3. Split-ring type Commutator. It consists of two halves (R₁ and R₂) of a metallic ring. The two ends of the armature coil are connected to these two halves of the ring. Commutator reverses the direction of current in the armature coil.
4. Two carbon brushes B₁ and B₂ press against the commutator. These brushes act as the contacts between the commutator and the terminals of the battery.
5. A battery is connected across the carbon brushes. This battery supplies the current to the armature coil.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric cars, rolling mills, electric fans, hair dryers, mixers, blenders etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is

1. pushed into the coil,
2. withdrawn from inside the coil,
3. held stationary inside the coil ?

Name the phenomenon involved in the above cases.
(CBSE 2010, Term I, 2011, 2012, 2013, 2014)

Answer:

1. When a bar magnet is pushed into the coil, induced current flows through the coil due to the phenomenon of electromagnetic induction. This induced current is indicated by the deflection of the needle of the galvanometer as shown in figure (a).
2. When a bar magnet is withdrawn from inside the coil, again induced current flows through the coil due to the phenomenon of electromagnetic induction. In this case, the direction of induced current is opposite to the direction of the current in case (i) as shown in figure (b).
3. When the bar magnet is held stationary inside the coil, there is no change in magnetic field around the coil. Hence, no induced current flows through the coil. Therefore, galvanometer shows no deflection as shown in figure (c).
   Phenomenon involved in electromagnetic induction.

Question 14.
Two circular coils A and B are placed close to each other. If the current in the coil A is changed, will some current be induced in the coil B 1 Give reason. (CBSE 2010, 2012)
Answer:
When current in coil A is changed, a changing magnetic field is set up around it. This changing magnetic field also links with coil B and hence some current will be induced in coil B due to electromagnetic induction.

Question 15.
State the rule to determine the direction of a

1. magnetic field produced around a straight conductor carrying current,
2. force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
3. current induced in a coil due to its rotation in a magnetic field.
   (CBSE 2010, 2011, 2014)

Answer:

1. Right hand thumb rule,
2. Fleming’s left hand rule,
3. Fleming’s right hand rule.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes ?
Answer:
An electric device used to convert mechanical energy (kinetic energy) into electrical energy (electricity) is called an electric generator.
Principle: Electric generator works on the principle of electromagnetic induction. When the coil of electric generator rotates in a magnetic field, induced current flows in the circuit connected with the coil.
types of electric generator

1. AC generator
2. DC generator

AC generator: AC generator converts mechanical energy into electrical energy in the form of alternating current or AC.
DC generator: DC generator converts mechanical energy into electrical energy in the form of direct current or DC.                                                AC Generator Construction : The main components of AC generator are (Figure 33) :

1. Armature : Armature coil (ABCD) consists of large number of turns of insulated copper wire wound over a soft iron core.
2. Strong field magnet : A strong permanent magnet or an electromagnet whose poles (N and S) are cylindrical in shape is a field magnet. The armature coil rotates between the pole pieces of the field magnet. The uniform magnetic field provided by the field magnet is perpendicular to the axis of rotation of the coil.
3. Slip Rings : The two ends of the armature coil are connected to two brass slip rings R₁ and R₂. These rings rotate along with the armature coil. Rings R₁ and R₂ are at different heights.
4. Brushes : Two carbon brushes (B₁ and B₂), are pressed against the slip rings. The brushes are fixed while slip rings rotate along with the” armature. These brushes are connected to the external circuit across which the output is obtained.

Working : When the armature coil ABCD rotates in the magnetic field provided by the strong field magnet, it cuts the magnetic field lines. Thus, the changing magnetic field produces induced current in the coil. The direction  of the induced current in the coil is determined by the Fleming’s right hand rule.
The current flows out through the brush B₁ in one direction in the first half of the revolution and through the brush B₂ in the next half revolution in the reverse direction. This process is repeated. Therefore, induced current produced is of alternating nature. Such a current is called alternating current.
DC generator or Dynamo Construction:

1. Armature coil. It consists of large number of turns of insulated copper wire wound on iron core in the form of a rectangle coil. Rectangle coil ABCD shown in figure 34 is an armature coil.
   
2. Strong field magnet. Armature coil is placed between two pole pieces (N and S poles) of a strong magnet. This magnet provides a strong magnetic field.
3. Split-ring Type Commutator. It consists of two halves (R₁ and R₂) of a metallic ring. The two ends of the armature coil are connected to these two halves of the ring.
4. Two carbon brushes B₁ and B₂ press against the commutator.
5. The output is shown by the glowing bulb connected across the carbon brushes.

Working of d.c. generator: When the coil of d.c. generator rotates in the magnetic field, induced potential difference is produced in the coil. This induced potential difference gives rise to the flow of current through the bulb and hence the bulb glows.
In d.c. generator, the flow of current in the circuit is in the same direction as long as the coil rotates in the magnetic field. This is because one brush is always in contact with the arm of the armature moving up and the other brush is in contact with the arm of the armature moving downward in the magnetic field.
Note: AC generator can be converted into DC generator by replacing slip rings used in AC generator by a split ring type commutator.

Question 17.
When does an electric short circuit occur ? (CBSE 2013)
Answer:
When live wire and neutral wire touch each other (i.e. come in direct contact.), the resistance of the circuit becomes small and hence large amount of current flows through it. As a result, large amount of heat is produced and the circuit catches fire.

Question 18.
What is the function of an earth wire ? Why is it necessary to earth metallic casings of electric appliances ? (CBSE 2010, 2011, 2012, 2013, 2015)
Answer:
Earth wire acts as a safety measure. When the live wire touches the metallic casing of an electric appliance, the electric current flows from the casing of the appliance to the earth through the copper wire. As the earth offers very’ low or almost no resistance to the flow of current, so large current passes through the copper wire instead of human body. This large current heats the circuit and hence the fuse in the circuit melts. As a result of this, the circuit is switched off automatically and hence the electric appliance is saved from burning and the human body suffers no electric shock.

NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

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NCERT Solutions for Class 10 Science Chapter 12 Electricity

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 12 Electricity. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 12 – Electricity solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 12 – Electricity Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register for our free webinar class with best Science tutor in India.

NCERT Questions

In Text Questions

Question 1.
What does an electric circuit mean ? (CBSE 2011, 2013, 2014)
Answer:
An electric circuit is a closed conducting path containing a source of potential difference or electric energy (i.e. a cell or battery) and a device or element utilizing the electric energy.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Define the unit of current. (CBSE 2008)
Answer:
Unit of electric current is ampere. Electric current in a conductor is said to be 1 A if 1 coulomb charge flows through the cross-section of the conductor in 1 second.

Question 3.
Calculate the number of electrons consisting one coulomb of charge. (CBSE 2015)
Answer:

Question 4.
Name a device that helps to maintain a potential difference across a conductor. (CBSE 2008)
Answer:
A cell or battery.

Question 5.
What is meant by saying that a potential difference between two points is 1 V ?
[CBSE (Delhi) 2008 ; CBSE (All India) 2008, 2010 Term I]
 Or
Define the term “Volt”. (CBSE 2009, 2013)
Answer:
Potential difference between two points is 1 V if 1 joule work is done in moving 1 coulomb charge from one point to another point.

Question 6.
How much energy is given to each coulomb of charge passing through a 6 V battery ?
Answer:
Energy = Charge x Potential difference =1 C X 6 V = 6 J.

Question 7.
On what factors does the resistance of a conductor depend ? (CBSE 2009, 2011, 2012, 2013, 2014)
Answer:
Resistance of a conductor depends on

1. length (l) of the conductors,
2. area of corss-section of the conductor,
3. nature of the material of the conductor and
4. temperature of the conductor.

Question 8.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source Why ?
Answer:
Area of cross-section

therefore, resistance of thin wire is more than the resistance of thick wire. Hence, current in thick wire flows easily than in thin wire.
Let the resistance of an electrical component remains constant while the potential difference across the ends of the component decreases to half of its former value.

Question 9.
What change will occur with current through it ?
Answer:

Thus, current in the component becomes half of its former value.

Question 10.
Why are coil of electric toasters and electric irons made of an alloy rather than a pure metal ?
(CBSE 2011, 2012)
Answer:
This is because the resistivity of an alloy is more than the resistivity of a pure metal and hence more heat is produced in an alloy than in pure metal due to the flow of current. Moreover, alloy does not burn (or oxidise) easily even at higher temperature.

Question 11.
(a) Which among, iron and mercury is a better conductor ? (resistivity of iron = 10.0 x 10^-8 Ω m and resistivity of mercury = 94 x 10^-8 Ω m)
(b) Which material is the best conductor ?
Answer:
(a) A material whose resistivity is low is a good conductor of electricity. Therefore, iron is better conductor than mercury.
(b) Silver is the best conductor of electricity.

Question 12.
Draw a schematic diagram of a circuit consisting of a batteries of three of 2 V each, a 5 Ω resistor, 8 Ω resistor and a 12 Ω resistor and a plug key, all connected in series. (CBSE 2011)
Answer:

Question 13.
Redraw the circuit of question 12, putting an ammeter to measure the current through the resistor and a voltmeter to measure the potential difference across 12 Ω resistor. What would be the reading in the ammeter ?
Answer:

Question 14.
Judge the equivalent resistance when the following are connected in parallel :
(a) 1 Ω and 10⁶ Ω
(b) 1 Ω and 10³ Ω and 10⁶ Ω.
(CBSE 2013)
Answer:
(a) When resistors are connected in parallel, then equivalent resistance of the combination is less than the least resistance in the combination. Therefore, equivalent resistance of 1 Ω and 10⁶ Ω connected in parallel is approximately 1 Ω but less than 1 Ω.
(b) The equivalent resistance is approximately 1 Ω but less than 1 Ω.

Question 15.
An electric lamp of 100 W, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current through it ?
Answer:

Question 16.
What is
(a) highest
(b) lowest resistance that can be secured by combining four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω ? (CBSE 2010 Term I, 2012, 2013)
Answer:

Question 17.
Why does the connecting cord of an electric heater not glow while the heating element does ? [CBSE (Delhi) 2008, 2010, 2011, 2013]
Answer:
This is because resistance of cord of electric heater is less than the resistance of heating element. So more heat is produced in the heating element and less heat is produced in the cord. Due to more heat, heating element glows.

Question 18.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V. (CBSE 2012, 2013)
Answer:

Question 19.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 seconds. (CBSE 2010)
Answer:

Question 20.
What are the advantages of connecting electrical devices in parallel with a battery instead of connecting them in series ?
                                                                           Or
Give two advantages of connecting electrical devices in parallel with battery. (CBSE 2010 Term I)
                                                                           Or
In a house-hold electric circuit different appliances are connected in parallel to one another. Give two reasons. (CBSE Sample Papers, CBSE 2012)
                                                                          Or
Why is parallel arrangement used in domestic ? (NCERT Question Bank, CBSE 2015)
Answer:

1. If any one of the electric devices in parallel fuses, then the working of other devices will not be affected.
2. When different devices are connected in parallel, they draw the current as per their requirement and hence they work properly.

Question 21.
How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of
(a) 4 Ω
(b) 1 Ω ? (CBSE 2012)
Answer:
(a) We get 4 Ω resistance if 3 Ω and 6 Ω resistors are connected in parallel and this parallel combination is . connected in series with 2 Ω as shown in figure.

Question 22.
What determines the rate at which energy is delivered by an electric current ? [CBSE (All India) 2008]
Answer:
Electric power determines the rate at which energy is delivered by an electric current.

Question 23.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and energy consumed in 2 h.
[CBSE (All India) 2008]
Answer:
Power, P = VI = 220 x 5 = 1100 W
Energy consumed = Power x Time
= 1100 W x 2 h = 2200 Wh = 2.2 kWh.

NCERT Chapter End Exercises

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Answer:

∴ (d) is correct answer.

Question 2.
Which of the following terms does not represent electrical power in a circuit ?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer:
(b) is correct answer.

Question 3.
An electric bulb is rated as 220 V and 100 W. When it is operated on 110 V the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:

∴ (d) is the correct answer.

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in an electric circuit. The ratio of heat produced in series and parallel combinations would be
(a) 1 :2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1.
Answer:

(c) is the correct answer.

Question 5.
How is voltmeter connected in circuit to measure the potential difference between two points ? (CBSE 2011, 2012)
Answer:

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10^-8  Ω m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if diameter is doubled ?
Answer:

When D is doubled and length remains the same, resistance becomes 1/4 th of the original resistance.

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :

Plot a graph between V and I and calculate the resistance of that resistor. (CBSE Sample Paper 2017-18)
Answer:

Question 8.
When a 12 V battery is connected across an unknown resistance, there is a current of 2.5 mA in the circuit. Find the value of resistance of the resistor.
Answer:

Question 9.
A battery of 9 V is connected in series with resistors of 0.2  Ω, 0.3  Ω, 0.4  Ω, 0.5  Ω and 12  Ω respectively. How much current would flow through 12 Ω resistor ? (CBSE 2010 Term I)
Answer:

Question 10.
How many 176  Ω resistors (in parallel) are required to carry 5 A on 220 V line ?
Answer:

Practical Skills Based Questions (Two Marks Questions)

Question 1.
In,a voltmeter, there are 20 divisions between the 0 mark and 0.5 mark. Calculate the least count of the voltmeter.
(CBSE 2015)
Answer:
Least count of voltmeter= value of 1 division
Here, 20 divisions = 0.5 V

Thus, least count of voltmeter = 0.025 V.

Question 2.
To verify Ohm’s law in the laboratory. Name the following circuit components used. (CBSE 2015)

Answer:
(a) Variable resistance
(b) Battery
(c) Wire crossing without joining
(d) Rheostat or adjustable resistance.

Question 3.
A student prepared a circuit diagram to study the dependence of potential difference (V) on electric current (I) across a resistor.

In this circuit diagram X, Y, Z, P are represented for the components.
(a) X = ……………..
(b) Y = ……………..
(c) Z = ………………
(d) P = ………………
(CBSE 2015)
Answer:
X = Ammeter
Y = Resistor
Z = Voltmeter
P = Battery.

Question 4.
Least count of a voltmeter and an ammeter in given figures would respectively be : (CBSE 2015)

Answer:
Least count of a voltmeter or an ammeter is equal to the value of 1 division on the scale.
In a voltmeter, 5 divisions = IV
or 1 division = 1/5 V = 0.2 V
Thus, least count of voltmeter = 0.2 mA
In an ammeter, 5 division = 1 mA
or 1 division = 1/5 mA = 0.2 mA
Thus, least count of ammeter = 0.2 mA.

Question 5.
To verify Ohm’s law, a student drew a circuit diagram which is given below :

1. Name two components in this circuit diagram which are connected in series.
2. Name two components in this circuit diagram which are connected in parallel. (C. B. S. E. 2015)

Answer:

1. Key, rhoestat, battery, ammeter and resistor are connected in series.
2. Resistor of resistance R and voltmeter (V) are connected in parallel.

Question 6.
In the Ohm’s law experiment, it is advised to take out the key from the plug when the observations are not being taken. Why it is essential l (CBSE 2015)
Answer:
Ohm’s law is valid only if the resistance of a resistor remains constant ‘during the experiment. If the key from the plug is not taken out when the observations are not taken, then the resistance of the resistor increases due to heat produced as a result of continuous flow of current through it. Hence, Ohm’s law cannot be verified.

Question 7.
The rest positions of the pointers of a milliammeter and volt-meter not in use are as shown in fig. A. When a student uses these in his experiment, the reading of pointers are in positions shown in fig. B.

Calculate the corrected value of current and voltage in this experiment.  (CBSE 2015)
Answer:
Least count of milliammeter = 10mA/5 = 2 mA
Least count of voltmeter = 1V/5 = 0.2 V
Zero error of milliammeter = 2 x 2 mA = 4 mA
Zero error of voltmeter = 3 x 0.2 V = 0.6 V
Correct value of current = 38 mA + 4 mA = 42 mA
Correct value of voltmeter = 3.6V – 0.6 V = 3.0 V.

Question 8.
In an experiment to study the dependence of current on potential difference across a resistor, a student obtained a graph as shown

Calculate the value of resistance of the resistor. (CBSE 2015)
Answer:

Question 9.
In a given ammeter, a student sees that needle indicates 17 divisions in ammeter while performing an experiment to verify Ohm’s law. If ammeter has 10 divisions between 0 and 0.5 A, then what is the value corresponding to 17 divisions ?  (CBSE Sample Paper 2017-18)
Answer:
10 Div = 0.5 A
1 Div = 0.5 A/10 = 0.05 A
∴ 17 Div = 0.05 A x 17 = 0.85 A

Question 10.
While performing an experiment to verify Ohm’s law, what precautions are to be taken ?
Answer:

1. Connections should be tight.
2. The conductor used should be such that its resistance does not change much with increase in temperature.
3. The plug of the key must only be inserted while reading ammeter and voltmeter. There after, the plug of the key must be taken out to avoid heating of conductor with the continuous flow of current through it.

Question 11.
In a given voltmeter, a student sees that the needle indicates 12 divisions in voltmeter while performing an experiment to verify Ohm s law. If voltmeter has 10 divisions between 0 and 1.0 V, then what is the value corresponding to 12 divisions ?
Answer:

Question 12.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :

Plot a graph between V and I and calculate the resistance of the resistor. (CBSE Sample Paper 2017-18)
Answer:
Scale : Along x-axis, 1 div = 0.1 V
Along y-axis, 1 div = 0.1 A
Graph between V and I is shown in figure:

Question 13.
An ammeter has a range of 0—3 ampere and there are 30 divisions on the scale. Calculate the least count of the ammeter. (CBSE 2015)
Answer:
30 divisions = 3 A
1 division = 3 A/30 = 0.1 A
Thus, least count of ammeter = 0.1 A

NCERT Solutions for Class 10 Science Chapter 12 Electricity

Hope given NCERT Solutions for Class 10 Science Chapter 12 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 11 – Human Eye and Colourful World solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 11 – Human Eye and Colourful World Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1.
What is meant by power of accommodation of the eye ?
(CBSE Sample Paper 2010, 2012, 2014, 2015, 2016, 2017)
Answer:
The ability of the eye lens to adjust its focal length to see nearby and distant objects clearly.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision ?
Answer:

Question 3.
What is the far point and near point of the human eye with normal vision ? (CBSE 2011, 2012, 2014)
Answer:
The farthest position of an object from the human eye so that its sharp image is formed on the retina is at infinite distance from the eye.
The nearest position of an object from a human eye so that its sharp image is formed on the retina is at 25 cm from the eye.

Question 4.
A student has difficulty in reading the black board while sitting in the last row. What could be the defect the child is suffering from ? How can it be corrected ? (CBSE 2012)
Answer:
Near sightedness or myopia. This defect can be corrected by using a concave lens of suitable focal length.

NCERT Chapter End Exercises

Question 1.
Choose the correct option :
Human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia
(b) near sightedness
(c) accommodation
(d) far sightedness.
Answer:
(c).

Question 2.
Human eye forms the image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina.
Answer:
(d).

Question 3.
The least distance of distinct vision for a young adult with normal vision is about (CBSE 2012, Bihar Board 2012)
(a) 25 m
(b) 2.5 cm
(c) 25 cm
(d) 50 cm. (CBSE 2011)
Answer:
(c).

Question 4.
The change in focal length of an eye lens is caused by the action of
(a) pupil
(b) retina
(c) ciliary muscles
(d) iris. (CBSE 2011)
Answer:
(c).

Question 5.
A person needs a lens of power – 5.5 diopters for correcting distant vision. For correcting his near vision, he needs a lens of power +1.5 diopter. What is the focal length of the lens required for correcting
(i) distant vision and
(ii) near vision ?
Answer:

Question 6.
The far point of a myopic person is 150 cm in front the eye. What is the nature and power of the lens required to correct the problem ?
Answer:

The lens is concave lens.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm.
(CBSE 2011)
Answer:
For diagram,

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm ? (CBSE 2011)
Answer:
The nearest position of an object from a normal human eye so that its sharp image is formed on retina is 25 cm. If the object is placed at a distance less than 25 cm, then the blurred image of the object is formed on retina as the focal length of eye lens cannot be decreased below a certain limit. Hence, eye cannot see it clearly.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye ?
[CBSE (Delhi) 2008, 2011]
Answer:
The image distance remains the same in the eye because the eye has the ability to change the focal length of its lens to make the image always on the retina when the object distance increases from the eye.

Question 10.
Why do stars twinkle ? (CBSE 2011, 2012, 2015, 2016)
                                     Or
Explain with the help of a labelled diagram, the cause of twinkling of stars. (CBSE 2014)
Answer:
Twinkling of Stars:
Light emitted by distant stars (act as point sources of light) passes through the atmosphere of the earth before reaching our eyes. The atmosphere of the earth is not uniform but consists of many layers of different densities. The layers close to the surface of the earth are optically denser. As we go higher and higher, the density of layers and refractive index decreases progressively. As the light from a star enters the upper­most layer of the atmosphere, it bends towards the normal as it enters the next layer. This process continues till the light enters our eyes. So due to refraction of light, the apparent position of the star is different from the actual position of the star (Figure 13).

Question 11.
Explain why the planets do not twinkle. (CBSE 2011, 2012, 2015)
Answer:
Planets do not twinkle:
Planets are very close to the earth as compared to the stars. The planets act as extended sources of light. So the intensity of light we receive from the planets is very large. Therefore, the variation in the brightness of the planets is not detected. Hence, planets do not twinkle.

Question 12.
Why does the Sun appear reddish early in the morning ? (CBSE 2011, 2012)
Answer:
When sunlight enters the atmosphere of the earth, the atoms and molecules of different gases present in the atmosphere absorb this light. Then these atoms and molecules of the gases re-emit light in all directions. This process is known as scattering of light. The atoms or particles scattering light are known as scatters.
The intensity of scattered light is inversely proportional to the fourth power of the wavelength of incident light, if the size of the particles (say atoms or molecules) scattering the light is less than the wavelength of the incident light.
That is, intensity of scattered light,                I ∝ 1/λ⁴.
We know, wavelength of red light is greater than the wavelength of blue or violet light. Therefore, the intensity of scattered red light is less than the intensity of the scattered blue or violet light.
The blue colour of sky, greenish blue colour of sea water, red colour of sunset and sunrise and white colour of clouds are due to the scattering of sun light by the particles present in the atmosphere of the earth.

Question 13.
Why does the sky appear dark instead of blue to an astronaut ?
[CBSE (Delhi) 2008, 2011, 2012, CBSE (Foreign) 2016]
                                                     Or
What will the colour of the sky be for an astronaut staying in international space station orbiting the earth 1 Justify your answer giving reason. (CBSE 2014)
Answer:
The blue colour of sky is due to the scattering of sunlight. The scattering of sunlight in the atmosphere is due to the presence of atoms and molecules of gases, droplets and dust particles. When the astronaut is in space, then there is no atmosphere (or atoms and molecules of gases, droplets and dust particles) around him. Therefore, sunlight does not scatter and hence sky appears dark.

Practical Skills Based Questions ( 2 Marks)

Question 1.
Draw a path of Light ray passing through a prism.
Label angle of incidence and angle of deviation in the ray diagram. (CBSE Sample Paper 2017-18)
Answer:

Question 2.
Draw the path of light ray passing through a glass prism. Label angle of incidence, angle of deviation and angle of emergence.
Answer:

Question 3.
The path of a ray of light through a glass prism is shown below :

Write the names of the angles represented by 1, 3, 4 and 5 respectively.
Answer:
1. represents angle of incidence.
3. represents angle of prism.
4. represents angle of deviation
5. represents angle of emergence

Question 4.
The path of a ray of light through a glass prism is shown below :

Name the incident ray of light, refracted ray of light and emergent ray of light.
Answer:

1. represents incident ray of light
2. represents refracted ray of light
3. represents emergent ray of light

Question 5.
The path of a ray of light passing through a glass prism is shown below :

Name the angles X, Y, Z and O
Answer:
∠X = Incident angle
∠Y = Angle of prism
∠Z = Angle of emergence
∠O = Angle of deviation

Question 6.
A beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 on a screen as shown in figure.

Name the colours similar to the colour of

1. danger or stop signal light
2. core of hard boiled egg
3. colour of clear sky
4. solution of potassium permanganate

Answer:

1. 1 is red colour which corresponds to the colour of danger or stop signal light.
2. 3 is yellow colour which corresponds to the colour of hard boiled egg.
3. 5 is blue colour which corresponds to the colour of clear sky.
4. 7 is violet colour which corresponds to the colour of the solution of potassium permanganate.

NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World

Hope given NCERT Solutions for Class 10 Science Chapter 11 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 9 – Heredity and Evolution solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 9 – Heredity and Evolution Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same species, which trait is likely to have arisen earlier ?
Answer:
In asexually reproducing population, there is no reshuffling of traits. New traits do develop due to small inaccuracies produced during DNA copying. They will be in smaller proportion than the traits already present. Therefore, trait B which exists in 60% of population must have arisen earlier than the trait A which occurs in 10% of the population.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
How does creation of variations in a species promote survival ?
Answer:
A number of different types of variations develop in a population. All of them do not have survival value. However, some of them are pre-adaptations which can be beneficial under certain environmental conditions. For example, in a heat wave most of the bacteria will die but a few having pre-adaptation or variation to tolerate heat wave will survive and multiply. Actually selection of variants by different environmental factors constitutes the basis for evolution.

Question 3.
How do Mendel’s experiments show that traits may he dominant or recessive ?
(CCE 2012, CBSE Delhi 2016, 2017)
Answer:
Mendel crossed Garden Pea plants having contrasting visible traits, e.g., tall and dwarf, violet and white flowered.
In F₁ generation there were no halfway characteristics. A cross between pure tall and pure dwarf plants yielded only tall plants in F₁ generation. There were no medium height plants. When F₁ plants were self bred, the F₂ plants were not all tall plants. Instead, both tall and dwarf plants appeared in ratio of 3 : 1. It means that the trait for dwarfness was present in F₁ generation but was not expressed while the trait for tallness expressed itself.

The trait of tallness which expresses itself in the presence of its contrasting form is called dominant. The other trait of dwarfness which is unable to express its effect in the presence of its contrasting trait is known as recessive.

Question 4.
How do Mendel’s experiments show that traits are inherited independently ? (CCE 2012, CBSE A.I. 2016, Delhi 17)
Answer:
Independent inheritance of traits is proved by employing dihybrid crosses and obtaining dihybrid ratios. Mendel crossed pure breeding tall plants having round seeds (TTRR) with pure breeding short plants having wrinkled seeds (ttrr). The plants of F₁ generation were all tall and with rounded seeds (TtRr) indicating that the characteristics of tallness and round seededness were dominant. Self breeding of F₁ yielded plants in the ratio of 9 tall round seeded, 3 tall wrinkled seeded, 3 short round seeded and one short wrinkled seeded. Tall wrinkled seeded and short round seeded plants are new combinations which can develop only if the traits are inherited independently. If the two traits are considered individually, F₂ ratio would be same as for monohybrid crosses, i.e., 12 tall : 4 short, 12 round seeded : 4 wripkled seeded.

TR, Tr, rR, tr x TR, Tr, rR, tr Gametes 9 tall rounded : 3 tall wrinkled : 3 short rounded : 1 short wrinkled

Question 5.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits—blood group A or O, is dominant ? Why or why not ?
Answer:
No. The information is not enough to tell whether the trait of the blood group A (I^A) or blood group
0(I⁰) is dominant. Either can be possible. Each individual carries two alleles. A recessive trait appears only when the two alleles are similar.
Possibility I: Blood Group A is Dominant and O Recessive. The trait of blood group O can appear only when both the recessive alleles occur together as in mother and daughter (I⁰I⁰). A group father should carry both the alleles of A and O (I^AI⁰).
Possibility 2: Blood Group O is Dominant and A Recessive. In this case the father should carry the alleles of A(I^AI^A) while the mother can be homozygous or heterozygous (I⁰I⁰, I⁰I^A). The daughter will have one dominant alleles of 0(I⁰I^A).
As both the possibilities can occur, the given information is unable to tell whether allele for blood group A or O is dominant.

Question 6.
How is the sex of the child determined in human beings ? (CCE 2013)
Answer:
Sex of the child is determined by the gametes that fuse to form zygote which later grows into offspring. Human females (44 + XX) produce only one type of ova (22 + X). Human males (44 + XY) form two types of sperms, androsperms (22 + Y) and gynosperms (22 + X). Both are formed in equal number. It is a chance factor whether an androsperm or a gynosperm fuses with egg to form 44 + XY or 44 + XX child. A child that obtains an X-chromosome from father will be girl and the one who inherits a Y-chromosome will be boy.

Question 7.
Draw Fig. 4.12. What are the different ways in which individuals with a particular trait may increase in a population ?
Answer:
There are three different ways in which individuals with a particular trait can increase in a population.

1. Survival Value (Natural Selection): The trait has survival value. It is picked up by natural selection. Through differential reproduction, it increases in population, e.g., green colour in beetles instead of red providing camouflage in bushes against being picked up by crows.
2. Genetic Drift: There is seasonal or accidental decline in population. The survivors have certain combination of traits which increase in number with the increase in population. The traits may not give any extra benefit to population.
3. Food: Individuals with particular trait may have extra abundance of food in their environment. They will naturally increase in number.

Question 8.
Why are traits acquired during the life time of an individual not inherited ?
(CCE 2011, 2012, 2013, CBSE A.I. 2017)
Answer:
Acquired traits are structural, functional and behavioural changes that an individual develops during its life time due to a particular environment, disease, trauma, use and disuse, conditioning or learning. The traits are not passed on to DNA of germ cells. They remain restricted to somatic cells. They are destroyed with the death of the individual. Therefore, intelligence, experiences and structural changes acquired during life time of an individual cannot pass to the progeny. Weismann (1892) cut the tails of mice for 21 generations but a tail still developed in 22nd generation.

Question 9.
Why are the small number of surviving tigers a cause of worry from the point of view of genetics ?
Answer:
A small population is always at a risk of degeneration and extinction due to

1. Excessive inbreeding that brings about inbreeding depression or degeneration,
2. Fewer recombinations and variations which are otherwise essential for maintaining vitality and vigour of the species.
3. Lesser adaptability to changes in the environment,
4. Increased threat to survival due to poaching, habitat destruction and environmental change.

Question 10.
What factors could lead to the rise of a new species ? (CCE 2012, 2014, CBSE Foreign 2017)
Answer:

1. Absence of gene flow amongst sub-populations due to the presence of physical barriers, long distance, differences in habitats, environmental and climatic conditions.
2. Accumulation of different variations in the different sub-populations of the species.
3. Natural selection of particular traits in a particular environment.
4. Genetic Drift. Separation of a small population, changes in its allele frequency, new mutations and adaptations to new habitat.
5. Reproductive Isolation. Accumulation of different variations and genetic drift result in absence of interbreeding in the previous subpopulations of a species. This results in the formation of new species. e.g., Finches of Galapogos islands.

Question 11.
Will geographical isolation be a major factor in the spéciation of a self pollinating plant species ? Why or why not ?
(CCE 2012)
Answer:
No. Geographical isolation has little role in spéciation of self pollinating plant species because there is already no gene flow among members of the species. Pea or Wheat which is self pollinated (due to pollination in bud condition) is not affected by any type of isolation. However, self pollinated plants can accumulate variations due to mutations and other factors and form new species.

Question 12.
Will geographical isolation be a major factor in the spéciation of an organism that reproduces asexually ? Why or why not ?
(CCE 2012)
Answer:
Recombination of genes is absent in asexually reproducing organisms. Therefore, variations originating in them do not get diluted but spread to all the» subsequent generations. Geographical isolation, which helps in spéciation due to formation of a separate gene pool, has no role in spéciation of asexually reproducing organisms.

Question 13.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
(CBSE Foreign 2017)
Answer:

1. Closeness of species is determined by presence or absence of fundamental characteristics and correlated characters. Two species of bacteria are closely related as they possess fundamental similarities of occurrence of nucleoid (instead of nucleus), absence of membrane covered cell organelles and presence of 70 S ribosomes. Human beings are close to monkeys because they possess similar eukaryotic multicellular body with vertebrate characters, mammalian traits and primate characters.
2. These days DNA matching is undertaken to find out the degree of closeness of the species.

Question 14.
Can the wing of a butterfly and the wing of a bat be considered homologous ? Why or why not ?
Answer:
No. Wings of butterfly and bat are fundamentally different in their origin and structure. In butterfly they are integumentary outgrowths having hollow tubes. In bat they are modified fore limbs which are covered by skin. Such organs which have a different origin and basic structure but are functionally similar are called analogous organs.

Question 15.
What are fossils ? What do they tell us about the process of evolution ? (CCE 2013, CBSE Foreign 2017)
Answer:
Fossils are remains or impressions of the past organisms that are found in the rocks of the old ages. They are often called written documents of evolution because they directly indicate the presence of different types of organisms in different ages. The path of evolution is known by arranging the fossils in a proper sequence age-wise. The early fossils are of simple organisms. Later on different complex forms arose, flourished and died down. They were replaced by newer forms. Study of fossils can also indicate the evolutionary stages of organisms. For example, modern horse (Equus) arose from a fossil animal Eohippus that existed on earth 60 million years back as a 30 cm high small animal. It evolved into 60 cm high goat sized Mesohippus about 40 million years back. Mesohippus gave rise to Merychippus (16-18 million years back) that formed Pliohippus (100-120 cm high, 10 million years ago). The modern horse evolved only 0-5 million years ago from Pliohippus.

Question 16.
Why are human being who look so different from each other in terms of size, colour and looks said to belong to same species ? (CBSE A.I. 2009 C, CBSE Foreign 2017)
Answer:
Delimitation of a species is based on the presence of a common gene pool, free inbreeding and reproductive isolation. Differences in size, colour and looks are based on preponderance of specific alleles and their interactions with the environment. All human beings, despite presence of different races, belong to same species {Homo sapiens) because they share the same gene pool, can marry amongst themselves and produce fertile offspring.

Question 17.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzee have a better body design ? Why or why not ?
Answer:
A better body design is the one which has more complexity, more elaboration and more controls which gives the organism a better competitive edge over others. There is no doubt that out of the four (bacteria, spiders, fish and chimpanzee), chimpanzee has a more elaborate body design or organisation. However, since body design is meant for competitive survival in their environment, all the four organisms or for that all living organisms, have a good body design that is suited to their environment.

NCERT Chapter End Exercises

Question 1.
A Mendelian experiment consisted of breeding tall Pea plants bearing violet flowers with short Pea plants bearing white flowers. The progeny all bore violet flowers but almost half of them were short. This suggests that genetic make up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) T_tWW
(d) T_tWw.
Answer:
(c) T_tWW.

Question 2.
An example of homologous organ is
(a) Our arm and a dog’s fore leg
(b) Our teeth and elephant tusks
(c) Potato and runners of grass
(d) All the above.
Answer:
(d) All the above.

Question 3.
In evolutionary terms we have more in common with
(a) A Chinese school boy
(b) A chimpanzee
(c) A spider
(d) A bacterium.
Answer:
(a) A Chinese school boy.

Question 4.
A study found that children with light coloured eyes are likely to have parents with light coloured eyes. On this basis can we say anything about whether the light eye colour is dominant or recessive ? Why or why not ?
Answer:
No. We cannot say with certainity whether the light eye colour is dominant or recessive. But since both the parents as well as the children have light eye colour, the probability is that it is a recessive trait. A recessive trait appears only when an individual possesses both the recessive alleles. As the parents are pure for the trait, the children also possess the trait and are pure for the same. Had the light eye colour been a domination trait, the recessive dark colour trait will have the chance to segregate and appear in some of the children.

Question 5.
How are the two areas of study, evolution and classification, interlinked ? (CBSE A.I. 2016)
Answer:
Classification is based on similarities and differences amongst organisms. The more characteristics two species have in common, the more closely related they are. They must have evolved from a common ancestor. Similarly more differences mean different adaptations and divergence from common ancestor in the remote past.

Question 6.
Explain the terms analogous and homologous organs with examples. (CBSE A.I. 2008 C)
Answer:
Analogous Organs: They are organs which have similar appearance and function but are quite different in their origin, development and anatomy.
Examples: Wings of Butterfly (integumentary outgrowths) and bird (modified fore-limbs).
Homologous Organs: They are organs which have similar origin, similar development and similar internal structure but have different forms and functions.
Examples: Fore-limbs of Horse, human hand, flipper of whale, wing of bird or bat.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:

1. Survey the dog population in and around your locality. Find out the percentage of different colours.
2. Observe the lineages where same colour is present in both parents and offspring. There is possibility that in these lineages both the alleles of coat colour are similar.
3. Allow crossing between two lineages having different coat colours.
4. Find the colour of F₁ individuals. It is probably the dominant coat colour.
5. Cross the F₁ dogs with the one having the other probably recessive colour. Is the ratio similar to test cross, Le., 1 : 1 ?

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
Fossils are remains or impressions of past organisms that are found in the rocks. Fossils of lower strata belong to early periods while those of upper strata are of later periods. Arranging the fossils stratumwise will indicate the occurrence of different forms of life at different times. It is found that the early fossils generally belong to simple organisms. Complexity and elaboration increased gradually with evolution. Evolution has never been linear or straight. A number of variants or branches appeared, some of which were more complex while others were less complex.

1. Fossils indicate the path of evolution of different groups.
2. They can indicate the phylogeny of some organisms, e.g, Horse, Elephant.
3. Some fossils have characteristics intermediate between two groups,
   e.g., toothed bird Archaeopteryx. They indicate how one group has evolved from another.

Question 9.
What evidence do we have for the origin of life from inanimate matter ?
Answer:
Miller and Urey (1953) assembled an apparatus which had a spark chamber (for producing lightning), a flask for boiling and a condenser. They introduced a mixture of methane, ammonia, hydrogen and water into the apparatus. The gaseous mixture was exposed to electric discharges, boiling (800°C) and condensation with the temperature kept just below 100°C. The experiment was continued for a few days. At the end of one week, 15% of carbon (from methane) had been converted into simple organic compounds of amino acids, organic acids, sugars and nitrogen bases. It clearly proved that organic compounds or building blocks of life developed from inanimate matter in the remote past when the hot earth was cooling.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually reproducing organism?
Answer:
Variations arising during sexual reproduction occur due to

1. Chance separation of homologous chromosomes during gametogenesis.
2. Crossing over between homologous chromosomes.
3. Chance coming together of chromosomes during fertilisation,
4. Errors or mutations occurring during DNA replication.

Only the last method of variations is found in asexually reproducing organisms. Therefore, rate of appearance of variations is quite high in sexually reproducing organisms as compared to asexually reproducing organisms. Further, variations developed in sexually reproducing organisms are quite viable as most of them are due to reshuffling of genes. It is not so in asexually reproducing organisms. Here, most of the changes are harmful. They have a negative impact on evolution except when changing environment finds them useful. Because of the abundance and viability of variations, the rate of evolution is also high in sexually reproducing organisms.

Question 11.
How is equal genetic contribution of male and female parents ensured in the progeny ?
Answer:
Most organisms are diploid. Their genetic material consists of two sets of chromosomes. Gametes carry single set of chromosomes, i.e., they are haploid. Sexual reproduction involves the formation and fusion of two types of gametes, male and female. Male gamete brings one set of chromosomes from the male parent. Female gamete also brings one set of chromosomes from the female parent. When two gametes fuse during sexual reproduction, the normal diploid chromosome complement is restored. It consists of 50% chromosomes from male parent and 50% chromosomes from female parent. Therefore, both the parents contribute equal genetic material to the offspring through formation and fusion of gametes.

Question 12.
Only variations that confer an advantage to an individual will survive in a population. Do you agree with this statement ? Why or why not ?
Answer:
No. Alongwith advantageous variations, a number of indifferent variations remain in the populations. Only the disadvantageous variations which are either lethal or extremely harmful are eliminated. All other variations persist in the population. Many of them function as preadaptations.

Selection Type Questions

Alternate Response Type Questions
(True/False (T/F), Right (√)/Wrong (x), Yes/No)

Question 1.
Mendel studied science and mathematics at the university of Amsterdam.
Question 2.
Both the parents contribute DNA equally to the offspring.
Question 3.
A factor which shows its effect in the hybrid is called recessive.
Question 4.
Sex of the child is determined by the type of ovum provided by the mother.
Question 5.
A recessive trait can also be common as blood group O.
Question 6.
Dromaesaurs were the first to fly.
Question 7.
Attached ear lobe is recessive trait.
Question 8.
Charles Darwin discovered the law of independent assortment,

Matching Type Questions

Question 9.
Match the articles given in columns A and B (single matching) :

Column A	Column B
(a)    Planaria (b)    DNA (c)    Miller and Urey (d)    Darwin	(i) Molecular Biology (ii) 1953 (iii) Natural selection (iv) Rudimentary eyes

Question 10.
Match the contents of columns I, II and III (double matching) :

Column I	Column II	Column III
(a)    Sex (b)    Variations (c)    Genetic drift (d)    Trilobite	(i) Adaptive (ii) Male gamete (iii) Palaeozoic arthropod (iv) New Variations	p.   New species q.  Jointed appendages r.  Conception s.  Increase

Question 11.
What type of similarity, homologous (H) and analogous (A) occurs in the pairs of organs ?

Pair of Organs	Similarity
(i) Forlimb of Horse, Wing of Bird (ii) Wings of Bat and Butterfly (iii) Forelimbs of Bird and Bat (iv) Wings of Bird and Bat

Question 12.
Match stimulus with appropriate Response

Trait	Dominant A	Recessive B	Acquired C
(i) Free ear Lobe (ii) Skin Tanning (iii) Wrinkled Seeds in Pea

Fill In the Blanks

Question 13.The term genetics was coined by ……………
Question 14.Mendel chose ………….. characters in Pea for his experiments.
Question 15. Broccoli has been developed from ……………… cabbage through artificial selection.
Question 16. ……………… Speciation occurs in geographically separated populations.
Question 17. Fossils are written documents of ………………..

Answers:

NCERT Solutions for Class 10 Science Chapter 9 – Heredity and Evolution

Hope given NCERT Solutions for Class 10 Science Chapter 9 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 10 – Light Reflection and Refraction solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 10 – Light Reflection and Refraction Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1.
Define the principal focus of a concave mirror.
Answer:
A point on the principal axis where the parallel rays of light after reflecting from a concave mirror meet.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is focal length ? (Bihar Board 2012)
Answer:
Radius of curvature, R= 20 cm.

Question 3.
Name a mirror that can give an erect and magnified image of an object.
Answer:
A concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles ?
[CBSE (All India) 2007, 2011, 2012]
Answer:
This is because a convex mirror forms an erect and diminished (small in size) images of the objects behind the vehicle and hence the field of view behind the vehicle is increased.

Question 5.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
R = +32 cm. Therefore, f = R/2 = +32/2 = +16 cm.
Thus, focal length of the convex mirror = +16 cm.

Question 6.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located ?
Answer:
m – -3, But m = -v/u, so v = 3u
u = -10 cm
v = 3 (-10 cm) =-30 cm
Thus, the image is located at a distance of 30 cm to the left side of the concave mirror.

Question 7.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal ? Why ?
Answer:
The ray of light bends towards the normal because the speed of light decreases when it goes from air (rarer medium) into water (denser medium).

Question 8.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass ? The speed of light in vacuum is 3 x 10⁸ m s^-1 (CBSE 2011)
Answer:

Question 9.
You are given kerosene, turpentine and water. In which of these does the light travel faster ?
Answer:
We know, v = c/n. Refractive index (n) of water is 1.333, whereas refractive index of kerosene is 1.44 and that of turpentine is 1.47. As refractive index of water is least, so speed of light in water is more than in kerosene and turpentine. Hence, light travels faster in water.

Question 10.
The refractive index of diamond is 2.42. What is the meaning of this statement ? [CBSE (Delhi) 2008, 2012, 2013; Bihar Board 2012]
Answer:

Question 11.
Define 1 dioptre of power of a lens.
Answer:
Power = I/f (in m).
Power of a lens is 1 dioptre if focal length of the lens is 1 metre or 100 cm.

Question 12.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object ? Also, find the power of the lens.
Answer:

Question 13.
Find the power of concave lens of focal length 2m?
Answer:

NCERT Chapter End Exercises

Question 1.
Which one of the following materials cannot be tised to make a lens 1
(a) water
(b) glass
(c) plastic
(d) clay.
Answer:
(d). This is because clay is opaque (i.e. light cannot pass through it).

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object ?
(a) between the principal focus and the centre of curvature
(b) at the centre of curvature
(c) beyond the centre of curvature
(d) between the pole of the mirror and its principal focus.
Answer:
(d).

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object ? (Bihar Board 2012)
(a) at the principal focus of the lens
(b) at twice the focal length
(c) at infinity
(d) between the optical centre of the lens and its principal focus.
Answer:
(b).

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of — 15 cm. The mirror and the lens are likely to be
(a) both are concave
(b) both are convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex but the lens is concave.
Answer:
(a).

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane only
(b) concave only
(c) convex only
(d) either plane or convex.
Answer:
(d).

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary ?
(a) a convex lens of focal length 50 cm
(b) a concave lens of focal length 50 cm
(c) a convex lens of focal length 5 cm
(d) a concave lens of focal length 5 cm.
Answer:
(c). Magnifying power of a reading glass (Convex lens) = 1/f.

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror ? What is the nature of the image ? Is the image larger or smaller than the object ? Draw a ray diagram to show the image formation in this case.
Answer:
A concave mirror produces an erect image if the object is placed between the pole and the focus of the concave mirror. Thus, object may be placed at any position whose distance is less than 15 cm from the concave mirror. The image is virtual and erect. The image is larger than the object. For a ray diagram, see figure 24.

Question 8.
Name the type of mirror used in the following situations :
(a) head lights of a car
(b) side rear view mirror of a vehicle
(c) solar furnace.
Support your answer with reason. (CBSE 2012, 2013)
Answer:
(a) Concave mirror. When a bulb is placed at the focus of a concave mirror, then the beam of light from the bulb after reflection from the concave mirror goes as a parallel beam which lights up the front road.
(b) Convex mirror. Image formed by a convex mirror is erect and small in size. The field of view behind the vehicle is large.
(c) Concave mirror. Concave mirror focuses rays of light coming from the sun at its focus. So, the temperature at the focus is raised.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object ?
(CBSE 2015)
Answer:
A complete image of the object is formed as shown in figure.

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Image is real and inverted.
Answer:

Here, u = -25 m, f = 10 m

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. Hou> far is the object placed from the lens ? Draw the ray diagram.
Answer:

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image. (CBSE 2014)
Answer:

Since, m is positive, so the orientation of both object and image is same. Thus,image is erect and virtual.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean ? (CBSE 2011, 2014)
Answer:

It means, size of the image formed by plane mirror is equal to the size of the object. Positive sign with m tells that both object and image are erect.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size. (Similar CBSE 2011)
Answer:
h = 5 cm, u = -20 cm


Since h’ is positive, so image is erect and virtual.

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained ? Find the size and the nature of the image.
Answer:

Since h’ is negative, so image is inverted.

Question 16.
Find the focal length of a lens of power -2.0 D. What type of lens is this ?
Answer:

The lens is concave.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging ? (CBSE 2012)
Answer:

Since focal length is positive, so the lens is converging.

Practical Skills Based Questions (2 Marks)

Question 1.
A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again ? How will the magnification of the image be affected ? (CBSE 2015)
Answer:

When distance of object u increases, V must decrease so that focal length of the lens remains the same. Thus, screen on which sharp image of object is formed is moved towards the lens.
As the distance of object u increases, the size of the image decreases. Hence, magnification of the imag decreases.

As u increases, v decreases and hence m also decreases.

Question 2.
In the given incomplete ray diagram for image A’B’ of a convex lens, what is the position of object AB ? Also complete the ray of diagram? (CBSE 2015)

Answer:
The position of object AB is between F₁ and 2F₁

Question 3.
To find the image-distance for varying object-distances in case of a convex lens, a student obtains on a screen a sharp image of a bright object placed very far from the lens. After that he gradually moves the object towards the lens and each time focuses its image on the screen.
(a) In which direction towards or away from the lens, does he move the screen to focus the object ?
(b) What happens to the size of image does it increase or decrease ?
(c) What happens when he moves the object very close to the lens ?
Answer:

As distance of object decreases, distance of image V must increase so that focal length of lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases, therefore, magnification increases. In other words, size of the image increases.
(c) When object is very close to convex lens, virtual and magnified image is formed.

Question 4.
A student performed an experiment for the image formation by a convex lens at different positions of an object. If focal length of lens is 15 cm
Answer:

Question 5.
A student performed an experiment with convex lens and found the virtual image of an object. Find :
(a) position of the object.
(b) draw ray diagram for the above situation.
(CBSE 2015)
Answer:
(a) A convex lens forms a virtual image if the object lies between the optical centre and focus- of the convex lens. That is, the distance of the object from the less must be less than the focal length of the lens.
(b)

Question 6.
A student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that he moves gradually the flame towards the lens and each time focuses its image on the screen.
(A) In which direction does he move the lens to focus the flame on the screen 1
(BJ What happens to the size of the image of the a formed on the screen 1
(C) What difference is seen in the intensity (brightness) of the image of the a screen ?
(D) What is seen on the screen when the flame is very close (at about 5 cm) to the lens ? (CBSE 2017)
Answer:

As the distance of the candle flame (object) u decreases, distance of image u must increase so that the focal length of the lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases therefore, magnification increases, other words, size of the image increases.
(C) As the size of the image increases, so the intensity (brightness) of the image of the flame on the screen decreases.
(D) When the flame is at about 5 cm, which is less than the focal length (10 cm) of the lens.
In this case, magnified, erect and virtual image of the flame is formed. Since, virtual image cannot be obtained on the screen, so nothing is seen on the screen.

Question 7.
A student places a candle flame at a distance of about 60 cm from a convex lens of focal length 10 cm and focuses the image of the flame on a screen. After that he gradually moves the flame towards the lens and each time focuses the image on the screen.
(A) In which direction-toward or away from the lens, does he move the screen to focus the image ?
(B) How does the size of the image change ?
(C) How does the intensity of the image change as the flame moves towards the lens ?
(D) Approximately for what distance between the flame and the lens, the, image formed on the screen is inverted and of the same size ? (CBSE 2017)
Answer:

As the distance of the candle flame (object) u decreases, distance of image u must increase so that the focal length of the lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases therefore, magnification increases, other words, size of the image increases.
(C) As the size of the image increases, so the intensity (brightness) of the image of the flame on the screen decreases.
(D) When object is placed at 2F, a real and inverted image of same size as that of the object is formed at 2F on the other side of the lens. Therefore, in this case, the distance between flame and the lens is approximately 20 cm so that the image formed on the screen is inverted and of the same size.

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

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NCERT Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 8 – How do Organisms Reproduce? solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 8 – How do Organisms Reproduce? Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
What is the importance of DNA copying in reproduction ? (CCE 2011, 2015)
Answer:
DNA carries hereditary information not only for controlling cellular functions but also all the structural and functional traits of organism. It is because of the latter that single celled zygote is able to form the whole multicellular organism. During reproduction there is formation of new cells which must carry the same amount and type of hereditary information as present in the parent cell. This is accomplished by DNA copying, which occurs prior to each cell division. DNA copying is not error proof. Errors give rise to variations.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Why are variations beneficial to the species but not necessarily for the individual ? (CCE 2011, 2012)
Answer:
Many of the variations are pre-adaptations which have no immediate benefit to the individuals. However, they remain in the population. Whenever, environment undergoes a drastic change, the pre-adaptations present in some members of the population allow the latter to survive, grow and regain its former size. Therefore, it is not necessary that variations are beneficial to individuals developing them but can prove useful to the species.

Question 3.
How does binary fission differ from multiple fission ?
Answer:

Binary Fission	Multiple Fission
1. Products. It gives rise to two individuals.	It forms several (more than two) individuals.
2. Conditions. Binary fission occurs under favourable conditions.	Multiple fission occurs both under favourable and unfavourable conditions.
3. Nucleus. Nucleus of the parent cell divides only once to form two daughters.	Nucleus of the parent undergoes repeated divisions to form a number of daughter nuclei.
4. Cytoplasm. Cytoplasm undergoes cleavage after each nuclear division.	Cytoplasm does not divide after every nuclear division.
5. Residue. No part of the parent body is left unused.	A part of the body, covering and residual cytoplasm, is left behind.
Examples. Amoeba, Paramecium.	Examples. Plasmodium, Amoeba (encysted).

Question 4.
How will an organism be benefited if it reproduces through spores ? (CCE 2011, 2012)
Answer:
Sporulation or spore formation is a method of asexual reproduction where each individual produces a number of spores. On germination each spore forms a new individual, e.g., Rhizopus.

1. All the daughters formed through spores are genetically similar.
2. Spores are a means of dispersal. They help in spreading the organism far and wide.
3. Spores can also function as a means of perennation or passage through unfavourable conditions.

Question 5.
Can you think of reasons why more complex organism cannot give rise to new individuals through regeneration.
Answer:
Regeneration is the ability of an organism to replace lost or injured parts so as to form the whole individual from an incomplete form or fragment by remodelling and growth of somatic cells through dedifferentiation, division, morphogenesis and redifferentiation. The ability for regenerative multiplication is present in simpler organisms because most of their cells can undergo dedifferentiation. However, it is limited to certain cells in complex organisms.

1. The stem cells of complex organisms can form lost tissues and organs but not the complete individual as the highly differentiated tissues and organs do not allow this.
2. In complex organisms regeneration is under neurohormonal control. Fragments do not have nervous or hormonal stimulus to grow into complete organisms.

Question 6.
Why is vegetative propagation practised for growing some types of organisms ?
Answer:
Vegetative propagation is practised in a number of horticulturally and economically important plants because it is advantageous.
Advantages:

1. Seedless Plants. Vegetative propagation is the only known method of multiplication of seedless plants, g., Banana, Sugarcane, Pineapple, Jasmine, some varieties of Orange, Rose.
2. Uniform Yield. Seeds and fruits are of uniform quality, size, taste and aroma.
3. Genetic Uniformity. Vegetative propagation gives a genetically uniform population.
4. Good Qualities. Good qualities of a variety can be maintained indefinitely.
5. Survival Rate. Survival rate of the daughters is nearly 100% while in case of seed grown plants, it is 10%.
6. Quicker Method. Vegetatively reproduced plants bear flowers and fruits earlier than the plants raised through seeds. Potato requires only three months for forming a new crop if raised from tubers. It takes 15 months if raised from seeds.
7. Introduction in New Areas. In areas where seed germination fails to form mature plants, vegetative reproduction can help in establishing the plants.

Question 7.
Why is DNA copying an essential part of the process of reproduction ?
Answer:
Cell multiplication is essential for reproduction either as a means of multiplication in unicellular organisms or as a means of development of multicellular body from a single celled zygote. Cell multiplication cannot occur without DNA replication or DNA copying because each new cell must carry the full DNA complement.

Question 8.
How is the process of pollination different from fertilization ?
Answer:

Pollination	Fertilization
1.    Definition: It is transfer of pollen grains from anther to the stigma of a flower. 2.    Step: Pollination precedes fertilization. 3.    Purpose: It carries the male gamete producing pollen grains to the female sex organ. 4. Process: Pollination is a physical process. 5. Occurrence: It occurs only in seed plants.	It is the fusion of male and female gametes. Fertilization occurs only after pollination when the pollen grain has germinated and male gametes are carried into ovule. It actually brings about fusion of gametes. Fertilization is a physico-chemical (biological) process. It occurs in both plants and animals of various types.

Question 9.
What is the role of seminal vesicles and the prostate gland ? (CGE 2011, 2012)
Answer:
Seminal Vesicles: They secrete 60-70% of semen plasma that is alkaline and viscous having fructose (for nourishing the sperms), fibrinogen, proteins and prostaglandins. Prostaglandins cause movements in the genital tract of the female. Sperms are also activated by secretion of seminal vesicles.
Prostate Gland: It produces 20-30% of semen plasma. The secretion is alkaline and viscous. It has clotting enzyme and chemical essential for sperm activity.

Question 10.
What are the changes seen in girls at the time of puberty ?
Answer:

1. Breast size begins to increase. There is darkening of skin of nipples below the tips of breasts.
2. Menarche or beginning of menstruation.
3. Broadening of pelvis,
4. Deposition of fat in face, buttocks and thighs,
5. Increased vasculature of skin and hence increased warmth of skin,
6. Rounding of body ccthtours.
7. High pitched voice,
8. Slow growth of ovaries, fallopian tubes, uterus, vagina, enlargement of labia, etc.

Question 11.
How does the embryo get nourishment inside the mother’s body ? (CCE 2012)
Answer:
Embryo gets nourishment from mother’s body with the help of placenta through a cord called umbilical cord. Placenta contains many finger-like villi from the chorion covering of the embryo. They occur in contact with blood sinuses of the mother present in the endometrial lining of uterus. All nutrients (glucose, amino acids, vitamins, etc.) diffuse from mother’s blood into villi and from there to embryo through the umbilical cord.

Question 12.
If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases ?
Answer:
No, sexually transmitted diseases occur due to fluid to fluid contact that takes place in the vagina.

NCERT Chapter End Exercises

Question 1.
Asexual reproduction takes place through budding in
(A) Amoeba
(B) Yeast
(C) Plasmodium
(D) Leishmania.
Answer:
(B).

Question 2.
Which of the following is not a part of the female reproductive system in human beings.
(A) Ovary
(B) Uterus
(C) Vas deferens
(D) Fallopian tube.
Answer:
(C).

Question 3.
The anther contains
(A) Sepals
(B) Ovules
(C) Carpel
(D) Pollen grains.
Answer:
(D).

Question 4.
What are the advantages of sexual reproduction over asexual reproduction ?
Answer:
Asexual reproduction is monoparental, with no gametes, no meiosis and very little variations. Sexual reproduction is generally biparental involving fusion of gametes, meiosis and lot of variations.

Question 5.
What are the functions performed by testis in human beings ? (CCE 2011)
Answer:

1. Formation of sperms from germinal cells found in seminiferous tubules.
2. Secretion of hormone testosterone by Leydig cells. Testosterone induces secondary sexual characters at puberty. It helps in maintenance and functioning of secondary sex organs.

Question 6.
Why does menstruation occur ?
Answer:
Menstruation occurs in response to low level of estrogen and progesterone hormones which causes constriction of blood vessels in uterine wall, stoppage of nourishment to overgrown endometrium that sloughs off, passing out broken mucosal membrane, blood and mucus.

Question 7.
Draw a labelled diagram of L.S. flower.
Answer:

Question 8.
What are the different methods of contraception ?
Answer:

1. Mechanical Barriers like condoms, cervical cap, diaphragm.
2. Oral Contraceptives or oral pills like Mala D, Saheli
3. Intrauterine Contraceptive Devices (IUCD) like loop, bow, Cu-T.
4. Surgical Methods like vasectomy in males and tubectomy in females.

Question 9.
How are the modes for reproduction different in unicellular and multicellular organisms ?
Answer:

Unicellular Organisms	Multicellular Organisms
1. Reproductive Cell. The same cell which functions as the body of the organism also gets transformed into reproductive cell.	Specific cells take part in reproduction.
2. Technique. Techniques of reproduction are simple.	Techniques of reproduction are commonly complex.
3. Asexual Reproduction. It occurs through fission, budding and spore formation.	It occurs by several methods like fragmentation, regeneration, budding, spore formation, vegetative reproduction, etc.
4. Sex Organs. No special sex cell or sex organ is present.	They are present.
5. Sexual Reproduction. It occurs through isogamy to heterogamy.	It is commonly oogamous.

Question 10.
How does reproduction help in providing stability to population of species ?
Answer:

1. Replication of DNA.
2. Growth and differentiation of cellular machinery.
3. Cell division. It is mode of reproduction in single celled organisms,
4. Development of special reproductive structures and formation of new individuals.
5. Continued replication of DNA, growth and cell division, formation of tissues, organs, etc. and maturation into a multicellular organism.

Question 11.
What could be the reasons for adopting contraceptive methods ?
Answer:

1. Enjoying a good reproductive health.
2. Protecting from sexually transmitted diseases.
3. Restricting the number of children.
4. Spacing the birth of children so as to properly look after them, provide them proper education without depleting the resources of the family.
5. Controlling population.

Selection Type Questions

Alternate Response Type Questions
(True/False (T/F), Right(√)/Wrong (x), Yes/No)

Question 1.
Basic event in reproduction is creation of DNA copy.
Question 2.
Plasmodium multiplies by binary fission.
Question 3.
Bryophyllum propagates through spore formation.
Question 4.
Copper-T is a contraceptive device used by women.
Question 5.
Hibiscus has unisexual flowers.
Question 6.
At the time of birth a girl baby has thousands of immature eggs.
Question 7.
Ovulation occurs in reproductively active females roughly in the middle of menstrual cycle.
Question 8.
Sperms mature at a temperature higher than that of human body.

Matching Type Questions

Question 9.
Match the articles given in columns A and B (single matching)

Question 10.
Match the contents of columns I, II and III (double matching) :

Question 11.
List the type of reproduction (A-asexual, V-vegetative, S-sexual) in the following organisms (Key or check list items)

Question 12.
Match each stimulus with appropriate response :

Fill In the Blanks

Question 13. ………………….. help in survival of the species in changing environment.
Question 14. ………………….. is common method of multiplication of Yeast and Hydra.
Question 15. Reproductive tissues begin to …………………….. when rate of general body growth begins to ………………… .
Question 16. Fallopian tubes are cut and ligated in ………………… .
Question 17. Bartholin’s glands are components of ……………………. reproductive system.

Answers:

NCERT Solutions for Class 10 Science Chapter 8 – How do Organisms Reproduce?

Hope given NCERT Solutions for Class 10 Science Chapter 8 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

In this chapter, students will come to know about early attempts at the classification of elements, Newlands Law Of Octaves, Mendeleev’s Periodic Table, the modern periodic table, metallic and non-metallic properties. At the end students are supposed to attempt question and answer exercise, multiple choice questions and group activity.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 5 – Periodic Classification of Elements solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 5 – Periodic Classification of Elements Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

Periodic Classification of Elements Class 10 Science In Text Book Questions

Question 1.
Did Doberiener’s triads also exist in the columns of Newlands’ octaves ? Compare and find out.
Answer:
Yes, some of the Doberiener’s triads did exist in the columns of Newlands’ octaves. For example,

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
What were the limitations of Doberiener’s triads ?
Answer:
Doberiener was in a position to identify three triads. It could not apply to all the elements known at that time. Therefore, the classification was not so useful.

Question 3.
What were the limitations of Nawlands’ Law of Octaves ?
Answer:

1. Actually this classification was successful only upto the element calcium. After that, every eighth element did not possess the same properties as by the element lying above it in the same group. For example, the elements cobalt (Co) and nickel (Ni) placed below chlorine had different properties. Same was the case with copper (Cu) placed after potassium in the same group.
2. Newland committed another mistake. He placed two elements in the same slot in a particular group. For example, Co and Ni in the first group after chlorine. Similarly, elements cerium (Ce) and lanthanum (La) were placed after yitterium (Y) in the same group. Newland could not offer any explanation for such an arrangement.
3. Newland somehow thought that only 56 elements existed in nature and no more elements were likely to be discovered. But this belief ultimately proved to be wrong.
4. When noble gas elements were discovered at a later stage, their inclusion in the table disturbed the entire arrangement.

Question 4.
Use Mendeleev’s periodic table to predict the formulae for the oxides of the elements : K, C, Al, Si, Ba.
Answer:
Oxygen is a member of group VTA in Mendeleev’s periodic table. Its valency is 2. Similarly, the valencies of all the elements listed can be predicted from their respective groups. This can help in writing the formulae of their oxides.

1. Potassium (K) is a member of group IA. Its valency is 1. Therefore, the formula of its oxide is K₂O.
2. Carbon (C) is a member of group IVA. Its valency is 4. Therefore, the formula of its oxide is C₂O₄ or CO₂.
3. Aluminium (Al) belongs to groups IIIA and its valency is 3. The formula of the oxide of the element is Al₂O₃.
4. Silicon (Si) is present in group IVA after carbon. Its valency is also 4. The formula of its oxide is Si₂O₄ or SiO₂
5. Barium (Ba) belongs to group IIA and the valency of the element is 2. The formula of the oxide of the element is Ba₂O₂ or BaO.

Question 5.
Besides gallium which two other elements have since been discovered that fill the gaps left by Mendeleev in creating his periodic table ?
Answer:
Two other elements are scandium (Sc) and germanium (Ge). In their gaps, the elements with names Eka-boron and Eka-silicon were placed.

Question 6.
What was the criteria used by Mendeleev in creating his periodic table ?
Answer:
Mendeleev used atomic masses of the elements as the criteria for creating his periodic table. In this table, the elements were arranged in order of increasing atomic masses.

Question 7.
Why do you think that the noble gases should be placed in a separate group ?
Answer:
In the Mendeleev’s periodic table, the elements have been arranged in the different groups on the basis of valency. For example, the elements placed in group I (IA and IB) have Valency equal to one. Same is the
case with the elements placed in other groups. Since the noble gas elements He, Ne, Ar, Kr, Xe, and Rn have zero valency, they could not find a place in Mendeleev’s periodic table. These have been placed in a separate group called zero group in the periodic table. Please note that the noble gas elements were not a part of the Mendeleev’s periodic table. They were added later on.

Question 8.
How could the Modern Periodic table remove various anomalies of Mendeleev’s periodic table ?
Answer:
We have already studied that the Mendeleev’s periodic table is based on the atomic masses of the elements whereas Modern Periodic Table takes into account their atomic numbers. Since the properties of the elements are linked with the electronic configuration of their atoms (i.e., atomic number), this means that Modern Periodic Table is better than the Mendeleev’s Periodic Table. The important advantages are listed.

1. The position of the elements in the periodic table are linked with their electronic configuration.
2. Each group is an independent group and the idea of sub-groups has been discarded.
3. One position for all the isotopes of an element is justified since the isotopes have the same atomic number. For example, the three isotopes of the element hydrogen e., protium, deuterium and tritium have atomic number one (Z = 1). Similarly, two isotopes of chlorine i.e. Cl-35 and Cl-37 are placed in the same slot since they have same atomic number (Z = 17).
4. The positions of certain elements which were earlier misfits in the Mendeleev’s periodic table  are now justified because it is based on   atomic numbers of the elements.  For example, Ar precedes K because its atomic number (18) is less than that of K (19).
5. It is quite easy to remember and reproduce.

Question 9.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice ?
Answer:
Magnesium (Mg) belongs to group 2 known as Alkaline Earth Family. The two other elements belonging to the same group are calcium (Ca) and strontium (Sr). The basis of choice is the electronic distribution in the valence shell of these elements. All of them have two electrons each.

	K	L	M	N	O
Mg (Z =12)	2	8	2	—	—
Ca (Z = 20)	2	8	8	2	—
Sr (Z = 38)	2	8	18	8	2

Question 10.
Name :
(a) three elements that have a single electron in their outermost shells.
(b) three elements that have two electrons in their outermost shells.
(c) three elements with filled outermost shells.
Answer:
(a) Lithium, sodium, potassium (Alkali metals present in group 1)
(b) Beryllium, magnesium, calcium (Alkaline earth metals present in group 2)
(c) Helium, neon, argon (Noble gases present in group 18).

Question 11.
(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements ?
(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common ?
Answer:
(a) The atoms of all these elements have one electron each in their valence shell. That is why, these elements are placed in the group 1 known as alkali metal group. The electronic configurations of these elements are given :

	K	L	M	N
Li (Z = 3)	2	1	—	—
Na (Z = 11)	2	8	1	—
K (Z = 19)	2	8	8	1

All the three elements evolve hydrogen gas on reacting with water
2 Li + 2 H₂O ———-> 2 LiOH + H₂
2 Na + 2 H₂O ————> 2 NaOH+ H₂
2 K + 2 H₂O ————> 2 KOH + H₂
Apart from this, all the elements happen to be the first elements of their respective periods. For example,

• Second period starts from lithium (Li)
• Third period starts from sodium (Na)
• Fourth period starts from potassium (K).

(b) Both these elements have completely filled shells. Helium (Z = 2) has two electrons in the only shell (K shell). The other element neon (Z = 10) has both K and L shells fully filled (2, 8). Because of the filled shells, the atoms of these elements do not have any desire to take part in chemical combination and they have been placed together in the same group known as group 18 or zero group.

Question 12.
In the modern periodic table, which are the metals among the first ten elements ?
Answer:
Metals among the first ten elements are lithium (Li) and beryllium (Be) . These are placed towards the left of the table.

Period	Group 1	Group 2	Group 13	Group 14	Group 15	Group 16
1 2	— —	— Be	— —	— —	— —	— —
3 4	— —	— —	— Ga	— Ge	— As	— Se

Question 13.
By considering their position in the periodic table, which one of the following elements would you expect to have maximum metallic characteristics ?
Answer:
Before identifying the metallic character from the list of the elements, we must remember two points :

• The metallic character of an element is related to the electron releasing tendency of its atoms. Greater the tendency, more will be the metallic character.
• In general, metallic character of the elements increases down the group and decreases along a period.
  With the help of the Modern Periodic Table, let us identify the group and period to which these elements

Since the metallic character increases down a group and decreases along a period, the obvious choice is between two elements. These are Be (beryllium) present in group 2 and Ga (gallium) present in group 13. Now, the size of Ga is very big as compared to that of Be. Actually, the element Ga has three shells since it belongs to period 4 while the element Be has only two shells as it is a member of period 2. This means that the element Ge has a greater electron releasing tendency of its atom as compared to the element Be. Therefore, Ge has more metallic character, rather maximum metallic character among the elements that are listed.

Periodic Classification of Elements Class 10 Science NCERT End Exercise

Question 1.
Which of the following statements is not correct about the trends while going from left to right across the periodic table ?
(a) The elements become less metallic in nature
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c). The atoms lose their valence electrons with difficulty and not easily. This is on account of the reason that

• nuclear charge increases from left to the right since the atomic number of the elements gradually increases.
• with the increase in nuclear charge, the force binding the electrons with the nucleus increases. Therefore, the atoms lose their valence electrons with difficulty.

Question 2.
Element X forms a chloride with the formula XCI2 which is a solid with high melting point. X would most likely to be in the same group of the periodic table as :
(a) Na
(b) Mg
(c) Al
(d) Si.
Answer:
(b). The formula of the chloride of the element is XCl₂. This means that the valency of the element X is 2 since chlorine is monovalent. The element with valency 2 is expected to be present in group 2 to which magnesium (Mg) belongs.

Question 3.
Which element has :
(a) two shells, both of which are completely filled with electrons ?
(b) the electronic configuration 2, 8, 2 ?
(c) a total of three shells with four electrons in the valence shell ?
(d) a total of two shells with three electrons in the valence shell ?
(e) twice as many electrons in the second shell as in the first shell ?
Answer:
(a) The elements with completely filled shells are noble gas elements and they belong to group 18. Since the element has two shells ; it must be present in second period and is neon (Ne) with electronic configuration 2, 8.
(b) The electronic configuration suggests that the element belongs to third period and second group. It is therefore, magnesium (Mg).
(c) The element with three shells is present in third- period. Since it* has four electrons in the valence
shell, it must belong to group 14 and is silicon (Si) with electronic configuration 2, 8, 4.
(d) The element with two shells is expected to be present in the second period. With three electrons in
the valence shell, it must belong to group 13 and is boron (B) with electronic configuration 2, 3.
(e) The element has only two shells. The first shell can have a maximum of two electrons. The second shell has four electrons which is twice the number of electrons present in the first shell. Therefore, the electronic configuration of element is 2, 4. It is carbon with atomic number (Z) equal to 6.

Question 4.
(a) Which property do all elements in the same column of the periodic table as boron have in common ?
(b) Which property do all elements in the same column of the periodic table as fluorine have in common ?
Answer:
(a) The element boron (B) is the first member of group (also called column) 13. It has three electrons in
the valence shell (2, 3). The other elements included in the same column are aluminium (Al),
gallium (Ga), indium (In) and thalium (Tl). They too have three electrons in the valence shell of their atoms. Just like boron, these elements also show a valency of 3 in their compounds.
(b) The element fluorine (F) is the first member of group (also called column) 17. It has seven electrons in the valence shell (2, 7). The other members present in the same group known as halogen family are chlorine (Cl), bromine (Br), iodine (I) and astatine (At). They have also seven electrons in the valence shell of their atoms. Like fluorine, they all show a valency of 1 in their compounds.

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element ?
(b) To which of the following elements would it be chemically similar ? (Atomic numbers are given in parentheses).
N(7), F(9), P(15), Ar(18).
Answer:
(a) The atomic number of the element is 17 (2 + 8 + 7 = 17).
(b) It would be chemically similar with fluorine (F) which has also 7 electrons in valence shell (2, 7)

Question 6.
The position of three elements A, B and C in the periodic table are shown below :

Group 16	Group 17
–	A
B	C

(a) State whether A is metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B ?
(d) Which type of ion, cation or anion will be formed by the element A ?
Answer:
(a) Group 17 represents halogen family. All the elements included in the family are Therefore, element A is a non-metal.
(b) Reactivity of non-metals is generally due to the electron accepting tendency of their atoms. Down the group, the atomic size increases. Therefore, the attraction of the nucleus for the outside electrons decreases. This means that down the group of non-metals, reactivity decreases. Thus, the element C is less reactive than the element A.
(c) Atomic size of the elements decreases along a period. The elements B and C are present in the same period. Since C is placed after B, the size of the element C is less than that of B.
(d) The elements A, as pointed out earlier is a non-metal which belongs to group 17. It has seven valence electrons (2, 8, 7). In order to have the configuration of the nearest noble gas element, it will take up one electron and change to anion i.e., A^– ion.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write their electronic configuration. Which of these will be more electronegative and why ? (CBSE 2012)
Answer:
The electronic configurations of the two elements are :
Nitrogen (Z = 7) 2, 5 ;
Phosphorus (Z = 15) 2, 8, 5
Since the size of nitrogen is small as compared to phosphorus, it has a greater tendency to take up electrons. It is therefore, more electronegative than phosphorus.

Question 8.
How does the electronic configuration of an atom relate to its position in the modern periodic table ?
(CBSE Delhi 2011)
Answer:
The modern periodic table is based upon atomic numbers of the elements. Since electronic configurations of the elements depend upon their atomic numbers, this means that the periodic table is based on the electronic configurations of the elements. For example, all the alkali metals have one electron each in their valence shell. These are placed in group 1. Similarly, the alkaline earth metals with two electrons in their valence shell are placed in group 2 and so on.

Question 9.
In the modern periodic table, calcium (Z = 20) is surrounded by the elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium ?
(CBSE All India 2011)
Answer:
Only those elements are placed in the same group in which the gaps of atomic numbers are : 8, 8, 18, 18, 32. If we look at the atomic numbers of the elements that are listed, it becomes clear that the elements with atomic numbers 12, 20 (Ca), 38 fit into this pattern. They are placed in the same group and have also similar physical and chemical properties.

Question 10.
Compare and contrast the arrangement of elements in Mendeleevs periodic table and the modern periodic table.
Answer:
The main points of distinction between Mendeleevs periodic table and Modern periodic table are as follows :

Mendeleev’s Periodic Table	Modern Periodic Table
1. It regards atomic masses of the elements as the basis of classification.	It regards atomic number of the elements as the basis of classification.
2. No separate positions or slots have been allotted to the isotopes of an element since they have different atomic masses.	There is no need for the separate slots for the isotopes since they have the same atomic numbers.
3. No justification is made for placing hydrogen in group IA along with alkali metals.	Justification has been made for placing hydrogen along with alkali metals in group 1 since both hydrogen and alkali metals have one valence electron.
4. Except for the elements in group VIII, the remaining groups have been divided with sub-groups A and B.	There are no sub-groups and all groups one independent in nature.
5. Position of certain elements based on their atomic masses are misfits. For example, the element cobalt (atomic mass = 58-9) has been its placed ahead of nickel (atomic mass = 58-7)	Modern periodic table is free from such anomalies. The element cobalt is placed before nickel since its atomic number (27) is less than that of nickel (28).
6. Electronic configurations and properties of the elements can not be predicted from their positions to the table.	Both electronic configuration and certain properties of the elements can be predicted from their positions in the periodic table.
7. It is not very systematic and is difficult to remember.	It is quite systematic and is easy to remember.

NCERT Solutions for Class 10 Science Chapter 5 – Periodic Classification of Elements

Hope given NCERT Solutions for Class 10 Science Chapter 5 helpful to you.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and its Compounds

In this chapter, students will learn about bonding in carbon –The Covalent Bond, versatile nature of Carbon, saturated and unsaturated carbon compounds, chemical properties of carbon compound, soaps and detergents.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 4 Carbon and its Compounds. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 4 – Carbon and Its Compounds solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 4 – Carbon and Its Compounds Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO₂ ?
Answer:
The atomic number (Z) for carbon is six and its electronic configuration is 2, 4. Carbon has four valence
electrons. Each oxygen atom (Z = 8) has six valence electrons (2, 6). In order to complete its octet, the carbon atom shares its four valence electrons with the four electrons of the two oxygen atoms as follows :

Thus, in carbon dioxide molecule, the carbon atom is linked to two oxygen atoms on both sides by two shared pairs of electrons resulting in double bonds on either sides. Both carbon and oxygen atoms complete their octet as a result of electron sharing.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur ?
Answer:
The atomic number (Z) of sulphur is sixteen and its electronic configuration is 2, 8, 6. The sulphur atom has six valence electrons. The chemical formula of sulphur molecule is S₈. Each sulphur atom is linked to similar atoms on either sides by single covalent bonds and thus, completes its octet. The molecule is in the form of a ring also represented by crown shape.
Ring structure of S₈ molecule Crown shape of S₈ molecule

Question 3.
How many structural isomers you can you draw for pentane ?
Answer:
Pentane (C₅H₁₂) has a skeleton of five carbon atoms. It can exist as a straight chain as well as two branched chains. There are three structural isomers for the hydrocarbon which is an alkane.

Question 4.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us ?
Answer:

1. Catenation: Carbon has the unique property of self linking which is known as catenation. In fact, any number of carbon atoms can be linked to one another by covalent bonds. This is on account of the stability of C—C bonds since the size of the carbon atom is quite small.
2. Linking of carbon with other atoms. Carbon is tetravalent in nature and can readily unite with atoms like hydrogen, oxygen, nitrogen, sulphur etc. by electron sharing.

Question 5.
What will be the formula and electron dot structure of cyclopentane ? (CBSE 2013)
Answer:
Cyclopentane is a cyclic compound with formula C₅H₁₂ The structure of the compound may be represented as :

Question 6.
Draw the structures of the following compounds :
(i) Ethanoic acid
(ii) Bromopentane
(iii) Butanone Are structural isomers possible for bromopentane ?
Answer:

Bromopentane has a chain of five carbon atoms. It can exist in a number of forms which are structural isomers.

The structural isomers (i), (ii) and (iii) which differ in the position of the Br atom are known as position isomers.
The structural isomers (iv), (v) and (vi) which differ in the arrangement of carbon atoms in the chain are called chain isomers.
In writing the IUPAC name, the name of prefix bromo is written before that of prefix methyl. In fact, alphabetical order is followed while naming the different prefixes.

Question 7.
How would you name the following compounds ?

Answer:
(i) Bromoethane
(ii) Hex-l-yne
(iii) Methanal

Question 8.
Why is the conversion of ethanol into ethanoic acid an oxidation reaction ?
Answer:
Ethanoic acid (CH₃COOH) has one oxygen atom more and two hydrogen atoms less than ethanol (C₂H₅OH). In general,

• Loss of hydrogen is known as oxidation.
• Gain of oxygen is known as oxidation.

Therefore, it is an oxidation reaction.

Question 9.
A mixture of ethyne and oxygen is used for welding. Can you tell why a mixture of ethyne and air is not used ?
Answer:
When ethyne is burnt in oxygen, large quantity of heat alongwith light is produced. The heat evolved can be used for gas welding which is usually carried to weld small broken pieces of articles made up of iron.

Air mainly contains a mixture of nitrogen (4 parts) and oxygen (1 part). As we know, nitrogen gas does not support combustion. This means that in air, only oxygen will help in the combustion of ethyne. Therefore, it is always better to use oxygen for the combustion of ethyne.

Question 10.
How would you distinguish experimentally between an alcohol and a carboxylic acid ?
Answer:
The distinction can be made by the following tests :

1. Dip a strip of blue litmus separately in both alcohol and carboxylic add taken in two glass tubes. The colour will change to red in the tube containing carboxylic acid and not in the tube which contains alcohol.
2. Add a small amount of solid sodium hydrogen carbonate (NaHCO₃) in both the tubes. A brisk effervescence accompanied by bubbles will be noticed in the tube containing carboxylic acid and not in the tube containing alcohol.

Question 11.
What are oxidising agents ?
Answer:
Oxidising agents are the substances which either their own or on reacting with another substance release
oxygen in order to carry oxidation reactions. The commonly used oxidising agents are : Ozone, bromine
water, a mixture of potassium dichromate and sulphuric acid or a mixture of potassium permanganate and sulphuric acid etc.

Question 12.
Would you be able to check if water is hard by using a detergent ?
Answer:
No, it is not possible. Actually detergents produce foam in any type of water ; whether hard or soft.
Therefore, a distinction between the two cannot be made. However, soaps can be used for this purpose.

Question 13.
People use a variety of methods to wash clothes. Usually after adding soap, they beat the clothes on a stone or beat them with a paddle, scrub with a brush or the mixture is agitated in a washing machine Why is this agitation necessary to get clean clothes ?
Answer:
The purpose of soap or detergent in washing is to reduce friction between oil drops carrying dirt particles and water so that they may mix with each other. All the methods that have been suggested loosen the bonds between the dust or oil particles and fabrics of clothes. The agitation helps in washing the clothes.

NCERT End Exercise

Question 1.
Ethane, with the molecular formula C₂H₆ has :
(a) G covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds
Answer:

(b). The molecule has seven covalent bonds.

Question 2.
Butanone is a four carbon compound with the functional group
(a) carboxylic acid
(b) aldehyde
(c) ketone
(d) alcohol.
Answer:
(c). The functional group is ketone (>C=0) also known as one.

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that
(a) the food is not cooked completely
(b) the fuel is not burning completely
(c) the fuel is wet
(d) the fuel is burning completely.
Answer:
(b). The fuel is not burning completely. The unburnt particles present in smoke blacken the vessel from outside.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH₃Cl.
Answer:
The molecule of chloromethane (CH₃Cl) consists of three elements i.e., carbon (Z = 6) hydrogen (Z = 1) and chlorine (Z = 17). Carbon atom has four valence electrons (2, 4) ; hydrogen has one (1) while chlorine has seven electrons in the valence shell (2, 8, 7). In order to complete its octet, carbon shares three valence electrons with three hydrogen atoms while one is shared with the electron of chlorine atom. The structure of covalent molecule may be written as follows :

Question 5.
Draw the electron dot structures for
(i) ethanoic acid
(ii) H₂S
(iii) propanone
(iv) F₂.
Answer:

Question 6.
What is homologous series ? Explain with an example.
Answer:
A series of similarly constituted compounds in which the members present have the same functional group, same chemical properties and any two successive members in a particular series differ in their molecular formula by —CH₂ group and molecular mass by 14 u.
For example, the boiling points of members in the family of alkanes follow the order :

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties ?
(CBSE Delhi 2011)

Answer:
Distinction based on physical properties :

1. Smell. Ethanol has a characteristic smell known as alcoholic smell which is pleasant. Ethanoic acid has vinegar like smell.
2. Boiling points. Boiling point of ethanol (351 K) is less than that of ethanoic acid (391 K).
3. Litmus test. Ethanol is neutral in nature and does not bring any change in the colour of litmus whether blue or red. Ethanoic acid is acidic and changes the colour of a blue litmus strip to red when dipped in it.

Distinction based on chemical properties :

1. Action with sodium hydrogen carbonate. On adding a small amount of sodium hydrogen carbonate to ethanoic acid, carbon dioxide gas is evolved with brisk effervescence. However, no such reaction is noticed in case of ethanol.
   CH₃COOH + NaHCO₃ ———> CH₃COONa + CO₂ + H₂O
2. Action with caustic alkalies. Ethanoic reacts with both sodium hydroxide (NaOH) and potassium hydroxide (KOH) to form corresponding salt and water. Ethanol fails to react with either of these.
   

Question 8.
Why does micelle formation take place when soap is added to water ? Will a micelle be formed in other solvents such as ethanol also ? (CBSE Delhi 2011)
Answer:
Soap may be represented by the formula RCOO^–Na⁺ where R is an alkyl group which represents long chain of carbon with fifteen or more atoms. Now, oil drops containing dirt particles and water donot mix. Soap helps in their mixing by reducing interfacial tension or friction. Actually it forms a sort of bridge between oil drops and water in which the alkyl portion (hydrophobic end) points towards oil drop while other portion COONa (hydrophilic end) is directed towards water. This is known as micelle formation. Thus, soap helps in the formation of a stable emulsion between oil and water. Ethanol and other similar solvents which are of organic nature, donot help in micelle formation because soap is soluble in them.

Question 9.
Why are carbon and its compounds used as fuels in most cases ?
Answer:
Carbon burns in oxygen or air to form carbon dioxide gas. The reaction is highly exothermic. That is why different forms of coal are used as fuels. The most important compounds of carbon are hydrocarbons. Just like carbon, hydrogen also readily burns in oxygen or air to form water producing heat. The hydrocarbon methane (CH₄) is a constituent of natural gas. Propane (C₃H₈) and butane (C₄H₁₀) are present in liquid petroleum gas (L.P.G.). Petrol and kerosene also contain different hydrocarbons. Therefore, these are used as fuels.

Question 10.
Explain the formation of scum when hard water is treated with soap. (CBSE Delhi 2011)
Answer:
Soap is basically sodium or potassium salt of higher fatty acids. Hard water contains in it Ca²⁺ and Mg²⁺ ions as their salts. When soap is added to hard water, the corresponding calcium and magnesium salts are formed. These are in the form of precipitates, also called ‘scum’.

Question 11.
What change will you observe by testing soap with litmus paper (blue or red) ?
Answer:
When soap is dissolved in water, the solution is alkaline in nature due to the formation of alkali NaOH or KOH. The solution changes the colour of red litmus to blue. However, the solution does not change the colour of blue litmus.

Question 12.
What is hydrogenation ? What is its industrial application ?
Answer:
The addition of hydrogen to unsaturated hydrocarbon in the presence of a metal catalyst is also known as catalytic hydrogenation.
The reaction is extremely useful in the hydrogenation of vegetable oils also called edible oils e.g. ground nut oil, cotton seed oil etc. These are also called cooking oils and are unsaturated in the sense that their molecules contain atleast one C=C bond in their structures. Upon passing hydrogen gas through oil in the presence of nickel catalyst, the double bond changes to single bond. As a result, the unsaturated oil changes to solid fat which is of saturated nature. Vegetable ghees such as Dalda, are of saturated nature and are formed by catalytic hydrogenation reaction.

Question 13.
Which of the listed hydrocarbons undergo addition reactions : C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄ ?
Answer:
In order that a hydrocarbon may undergo addition reaction, it must be unsaturated in nature. It must be either an alkene (C=C) with general formula C_nH_2n or an alkyne (C = C) with general formula C_nH_2n-2. Out of the list of the hydrocarbons given :

• C₃H₆ (Propene) is an alkene with C=C bond. It corresponds to general formula C_nH_2n (n = 3)
• C₂H₂ (Ethyne) is an alkyne with C = C bond. It corresponds to general formula C_nH_2n-2 (n = 2).

Both these hydrocarbons take part in addition reactions. For example, they react with hydrogen upon heating to 473 K in the presence of Nickel catalyst to form corresponding alkanes.

Question 14.
Give a test that can be used to differentiate between butter and cooking oil.
Answer:
Butter is saturated in nature while cooking oil is unsaturated. This means that cooking oil has atleast one C=C bond present in the constituting compounds while butter does not have any such bond. The distinction between them can be made by reacting with bromine water or bromine dissolved in carbon tetrachloride. Cooking oil will discharge the yellow colour of bromine while butter will not.

Question 15.
Explain the mechanism of cleansing action of soap.
Answer:
Cleansing action of Soaps and Detergents
Both soaps and detergents resemble in their cleansing action. They have two main parts. These are non—polar hydrocarbon chain which is water repellent or hydrophobic (phobic or phobia stands for repulsion or hatred) and the polar carboxyl group as its salt which is attracted towards water or is hydrophilic (philic stands for love or attraction). The former is called the tail of the molecule and the latter is regarded as the head.

In order to understand the cleansing action of both soaps and detergents, let us try to analyse how the clothes become dirty. They first become oily because of the perspiration coming out of the skin and also from the organic matter dispersed in the atmosphere. Dust particles stick to oil drops and the clothes become dirty. In order to wash these, they are dipped in water and soap or detergent is applied. In solution, it dissociates to give carboxylate ions (RCOO ) or sulphonate ions (RSO₃^–) and the cations (Na⁺). In the carboxylate ion, the alkyl portion which contains a long chain of hydrocarbons is a tail pointing towards the oil drops while the COO^– portion is the head directed towards water. In a detergent it is the SO₃^– portion which points towards water. This is quite evident from the figure where the solid circles (•) represent the polar groups and the wavy lines (^^^) represent the alkyl portions. The molecules of soap or detergent have a unique orientation in water. They actually form a cluster of molecules in which the hydrophobic or alkyl portion is in the interior while the ionic or polar portion is on the surface of the cluster as shown in the figure. This formation is known as micellear formation or simply micelle. Soaps or detergent thus help in forming a stable emulsion of oil and water by acting as a bridge between the two. The oil droplets alongwith the particles of the dirt get detached from the fibres of the clothes and pass into the emulsion. In this manner, the clothes become free from dust or dirt.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Hope given NCERT Solutions for Class 10 Science Chapter 4 helpful to you.

NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals

In this chapter, students will learn about the physical properties of metals, and non-metals, chemical properties of metals, how metals react with air, water, acid, other solutions, and metal salts, reactivity series.

Further students will come to know how metals and non-metals react, properties of ionic compounds, the occurrence of metals, extraction of metals, refining of metals, corrosion, prevention of corrosion. The chapter contains questions and answers exercise along with multiple choice questions.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 3 – Metals and Non-metals solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 3 – Metals and Non-metals Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 10 Science Chapter 3 NCERT Questions

NCERT Solutions for Class 10 Science Chapter 3 In Text Book Questions

Question 1.
Give example of a metal which
(a) is a liquid at room temperature
(b) can be easily cut with a knife
(c) is the best conductor of heat
(d) is the poorest conductor of heat.
Answer:
(a) Mercury
(b) Sodium
(c) Silver
(d) Lead.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Explain the meaning of malleable and ductile.
Answer:
Malleable: The property due to which a substance can be beaten into sheets is known as malleability. Metals are malleable in nature.
Ductile. The property due to which a substance can be drawn into wires is known as ductility. Metals are ductile in nature.

Question 3.
Why is sodium kept immersed in kerosene oil ? (CBSE 2011)
Answer:
Sodium reacts with air and water both. It is a highly reactive metal. When kept in open, it readily combines with oxygen present in air to form its oxide. Similarly, it reacts with water vapours or moisture to form sodium hydroxide.

In order to preserve sodium metal, we generally keep it under kerosene so that neither air nor moisture may come in its contact.

Question 4.
Write the equations for the reactions of
(a) iron with steam
(b) calcium with water
(c) potassium with water.
Answer:
(a) 3Fe(s) + 4H₂O (steam) ———–> Fe₃O₄(s) + 4H₂(g)
(b) Ca(s) + 2H₂O(aq) ———–> Ca(OH)₂(s) +H₂(g)
(c) 2K(s) + 2H₂O(aq) ———–> 2KOH(aq) + H₂(g)

Question 5.
Samples of four metals A, B, C and D were taken and were added to the following solutions one by one. The results obtained have been tabulated as follows :

Metal	Solution to which metal is added
	Iron(II) sulphate	Copper(II) sulphate	Zinc sulphate	Silver nitrate
A	No reaction	Displacement	—	—
B	Displacement	—	No reaction „	—
C	No reaction	No reaction	No reaction	Displacement
D	No reaction	No reaction	No reaction	No reaction

Use the table given above to answer the following questions :
(a) Which is the most reactive metal ?
(b) What would you observe when B is added to solution of copper(II) sulphate ?
(c) Arrange the metals A, B, C and D in order of increasing reactivity. (CBSE 2011)
Answer:
Based on the activity series, the relative position of the metals in which involved in solutions is : Zn > Fe > Cu > Ag. On the basis of the results given in the table .

• Metal A is more reactive than copper and less reactive than iron.
• Metal B is more reactive than iron and less reactive than zinc. –
• Metal C is only more reactive than silver and less reactive than other metals.
• Metal D is the least reactive in nature.

In the light of above information, we can conclude that
(a) Metal B is the most reactive.
(b) Since B is more reactive than iron, it is also more reactive than copper. This means that it would displace copper from copper(II) sulphate solution. The blue colour of solution will slowly fade.
(c) The decreasing order of reactivity of metals is: B>A>C>D.

Question 6.
Which gas is produced when a reactive metal reacts with dilute hydrochloric acid ? Write the chemical reaction when iron reacts with dilute H₂SO₄. (CBSE 2010)
Answer:
Hydrogen gas (H₂) is produced when a reactive metal reacts with dilute hydrochloric acid. Iron and dilute H₂SO₄ react as follows :
Fe(s) + H₂SO₄(aq) ————> FeSO₄(aq) + H₂(g)
Hydrogen gas is evolved in this reaction also.

Question 7.
What would you observe when zinc is added to a solution of iron (II) sulphate ? Write the chemical reaction that takes place. (CBSE 2010)
Answer:
The green colour of the solution would slowly disappear. Zinc would gradually dissolve and iron would get precipitated at the bottom of the beaker.

Question 8.
(i) Write electron-dot structures for sodium, magnesium and oxygen.
(ii) Show the formation of Na₂O and MgO by the transfer of electrons.
(iii) What are the ions present in these compounds ?
Answer:
(i)

(ii) Formation of sodium oxide (Na₂O)

Formation of magnesium oxide (MgO)

(iii) For answer, consult structures given above.

Question 9.
Why do ionic compounds have high melting points ? (CBSE 2014)
Answer:
In the formation of ionic compounds, positive ions (cations) and negative ions (anions) participate. These are closely packed and the ionic compounds exist as crystalline solids. They have strong inter ionic forces of attraction and have high melting and boiling points.

Question 10.
Define the following terms :

1. Minerals
2. Ores
3. Gangue.

Answer:

1. Minerals : These are the combined states of metals and non-metals present in earth’s curst.
2. Ores : The minerals from which metals can be conveniently and profitably extracted, are called ores.
3. Gangue : It represents the earthy impurities such as mud, sand and clay associated with the ore.

Question 11.
Name two metals which are formed in nature in free state.
Answer:
The metals are gold (Au) and platinum (Pt).

Question 12.
Which chemical process is used for obtaining a metal from its oxide ?
Answer:
The chemical process is known as reduction.

Question 13.
Metallic oxides of zinc, magnesium and copper were heated with the following metals. In which cases, will you find displacement reactions taking place ?

Metal	Zinc	Magnesium	Copper
Zinc oxide Magnesium oxide Copper oxide

Relative positions of these metals in the activity series are : Mg, Zn, Cu : In the light of this :
Answer:
Magnesium (Mg) will displace both zinc (Zn) and copper (Cu) from their oxides
Mg + ZnO ———-> MgO + Zn
Mg + CuO ———-> MgO + Cu
Zinc will displace copper from copper oxide.
Zn + CuO ———–> ZnO + Cu
Copper is least reactive and will not initiate displacement reaction.

Question 14.
Which metals do not corrode easily ?
Answer:
Metals such as gold (Au) and platinum (Pt) present at the bottom of the activity series do not corrode easily.

Question 15.
What are alloys ? (CBSE 2011)
Answer:
Alloys are the homogeneous mixture of two or more metals or even metals and non-metals.

NCERT Solutions for Class 10 Science Chapter 3 NCERT End Exercise

Question 1.
Which of the following will give displacement reactions ?
(a) NaCl solution and copper metal
(b) MgCl₂ solution and aluminium metal
(c) FeSO₄ solution and silver metal
(d) AgNO₃ solution and copper metal.
Answer:
(d). Only AgNO₃ solution will give displacement reaction with copper (Cu) because copper is placed above silver in the activity series.

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting ?
(a) applying grease
(b) applying paint
(c) applying a coating of zinc
(d) all the above.
Answer:
Theoretically all the three methods are helpful for preventing an iron frying pan from rusting. However, the constituents of both grease and paint are mostly organic compounds. They cannot withstand the heat and do not last. Therefore, applying a coating of zinc (galvanisation) is the best method. Option (c) is correct.

Question 3.
An element reacts with oxygen to give a compound with high melting point. This compound is also water soluble. The element is likely to be :
(a) Calcium
(b) Carbon
(c) Silicon
(d) Iron
Answer:
(a). Calcium (Ca) combines with oxygen to form calcium oxide (CaO) with very high melting point. CaO dissolves in water to form calcium hydroxide

Question 4.
Food cans are coated with tin and not with zinc because
(a) Zinc is costlier than tin
(b) Zinc has higher melting point than tin
(c) Zinc is more reactive than tin
(d) Zinc is less reactive than tin.
Answer:
(c). Zinc is more reactive than tin and reacts with organic acids present in food to form poisonous compounds. Since tin is placed below zinc in the activity series, it is less reactive and does not react with the organic acids. Therefore, (c) is the correct option.

Question 5.
You are given a hammer, a battery, a bulb, wires and a switch :
(a) Flow could you use them to distinguish between samples of metals and non metals ?
(b) Assess the usefulness of these tests to distinguish between metals and non metals.
Answer:
(a) With the help of hammer, convert both the metal and non-metal (solid) in the form of plates or rods. Metal will readily form these since they are malleable. Non-metals being brittle will break when struck with hammer. They will form plates with difficulty Now construct a cell in both the cases using these plates as electrodes and switch on the current. If the bulb glows, this means that the electrodes are of metals. In case, this does not glow, this means that the electrodes are of non-metals.
(b) From these tests, we may conclude that

1. Metals are malleable while non-metals are not.
2. Metals are good conductors of electricity while non-meals are not (graphite is an exception).

Question 6.
What are amphoteric oxides ? Give examples of two amphoteric oxides.
Answer:
These are the oxides which can act both as acids and bases. For example, aluminium oxide (Al₂O₃) and zinc oxide (ZnO). The amphoteric character of the two oxides are shown by the following reactions.

Question 7.
Name two metals which can displace hydrogen from dilute acids and two metals which can not do so.
Answer:
Sodium and calcium can displace hydrogen from dilute acids Copper and silver can not displace hydrogen from dilute acids.

Question 8.
In the electrolytic refining of metal M, name anode, cathode and electrolyte.
Answer:
Anode : Rod of the impure metal
Cathode : Rod of pure metal
Electrolyte : Aqueous solution of soluble salt of metal M.

Question 9.
Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it as shown in the figure.

What will be the action of gas on

1. dry litmus paper ?
2. moist litmus paper ?

Write a balanced chemical equation for the reaction taking place.
(CBSE 2011)
Answer:
The gas evolved upon heating sulphur powder on a spatula is sulphur dioxide

1. SO₂(g) has no action of dry litmus paper.
2. SO₂(g) dissolves in moisture (water) present in moist litmus paper to form sulphurous acid. In acidic solution, moist litmus paper will change to red.
   

Question 10.
State two ways to prevent the rusting of iron.
Answer:

1. By applying a coating of grease or paint on the surface of iron.
2. By depositing a layer of zinc on the surface of iron. The process is called galvanisation.

Question 11.
What types of oxides are formed when non-metals combine with oxygen ?
Answer:
The oxides are generally acidic in nature which means that when dissolved in water, their solutions change blue litmus to red. For example,

Question 12.
Give reasons for the following :
(a) Platinum, gold and silver are used to make jewellery.
(b) Sodium, potassium and lithium are stored under oil.
(c) Aluminium is a highly reactive metal but still used for making cooking utensils.
(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction. (CBSE 2013, 2014)
Answer:
(a) These metals placed at the bottom of the activity series are very little reactive in nature. Gold and
platinum are known as noble metals. They are not affected by air, water and by chemicals. Since they have bright lustre, jewellery can be made from these metals.
(b) There are reactive metals placed high in the activity series. In air, their surface gets tarnished due to presence oxygen, water vapours and carbon dioxide in air. With water, these react violently to evolve so much heat that is not possible to handle them. These metals are generally kept under kerosene which does not contain air and water.
(c) When exposed to air, the metal changes its oxide called aluminium oxide (Al₂CO₃). It gets deposited over the surface of the metal and forms a protective coating on the surface. Due to the presence of this layer, the metal becomes unreactive and can be used in making cooking utensils.
(d) Both carbonate and sulphide ores of metals cannot be directly reduced to metallic state. Flowever, metal oxides can be easily reduced with coke or other reducing agents. Both are therefore, converted into their respective oxides by calcination process (for carbonate ores) and by roasting process (for sulphide ores).
Metal oxides can be easily reduced to metallic form with coke (C) or any other suitable reducing agent. Therefore, carbonates and sulphides are converted to the oxide form by processes of calcination and roasting and are not directly reduced.

Question 13.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels. (CBSE 2014)
Answer:
Copper metal slowly reacts with water, carbon dioxide and oxygen present in air to form basic copper carbonate which is green in colour. Its layer slowly gets deposited on the surface of the metal. Now lemon

juice contains citric acid while tartaric acid is present is tamrind. Both these acids react with basic copper carbonate to form soluble salts such as copper acetate (with citric acid) and copper tartarate (with tartaric acid). The equations for the reactions are complicated and are not given. These salts are removed from the surface of the copper metal and the surface of the metal shines.

Question 14.
A man went door to door posing as a goldsmith. He promised to bring back the glitter on dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparked like new but their weight was reduced drastically. The lady got upset and after a futile argument, the man beat a hasty retreat. Can you predict the nature of the solution used by the impositer ?
Answer:
The man had actually used the solution of aqua regia (mixture of cone. HCl and cone. HNO₃ in the ratio of 3 : 1 by volume) which has dissolved gold forming soluble auric chloride (AuCl₃). Since gold actually reacted, there was a loss in weight of the gold bangles. With the removal of the dull layer of gold from the surface, there was original shine on the bangles. The chemical reactions have been

Question 15.
Give reason as to why copper is used to make hot water tanks and not steel (an alloy of iron).
Answer:
Copper is a better conductor of heat than steel which is an alloy of iron. Though copper is costlier than steel, it is used to make hot water tanks for storing hot water in preference to steel.

Question 16.
Differentiate between metals and non-metals on the basis of chemical properties.
For the distinction in the chemical characteristics,

Property	Metals	Non-metals
1. Nature of oxides	Oxides of the metals are generally basic in nature (Exception : ZnO and Al2O3 are amphoteric oxides).	Oxides of non-metals are mostly acidic in nature (Exception : CO and N2O are neutral oxides).
2. Electrochemical behaviour	Metals normally form cations by the loss of electrons. This means that these are electropositive in nature.	Non-metals normally form anions by the gain of electrons. This means that these are electronegative in nature.
3. Action with dilute acids	Active metals evolve hydrogen on reacting with dilute HCl and dilute H2SO4.	Non-metals do not react with dilute acids
4. Nature of compounds	The compounds of metals are mostly ionic in nature.	Compounds of non-metals are mostly covalent although there are many exceptions.
5. Oxidising and reducing nature	Metals act as reducing agents as their atoms lose electrons. For example, Na —— > Na+ + e–	Non-metals act as oxidising agents as their atoms accept electrons. For example, Cl + e–——–>Cl–

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NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals